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101. A force of 5 N acts on an object in the direction of displacement 3 m. What is the work done?
ⓐ. 8 J
ⓑ. 15 J
ⓒ. 3 J
ⓓ. 0 J
Correct Answer: 15 J
Explanation: \(W = \mathbf{F} \cdot \mathbf{d} = 5 \times 3 = 15\) J.
102. A force of 12 N is applied at an angle of \(60^\circ\) to the horizontal. If the object moves a distance of 8 m horizontally, what is the work done?
ⓐ. 24 J
ⓑ. 48 J
ⓒ. 96 J
ⓓ. 64 J
Correct Answer: 48 J
Explanation: The horizontal component of the force is \(12 \cos 60^\circ = 12 \times 0.5 = 6\) N. Therefore, \(W = 6 \times 8 = 48\) J.
103. A force of 20 N acts on an object that moves a distance of 4 m. If the force makes an angle of \(30^\circ\) with the direction of motion, what is the work done?
ⓐ. 80 J
ⓑ. 60 J
ⓒ. 40 J
ⓓ. 20 J
Correct Answer: 40 J
Explanation: \(W = \mathbf{F} \cdot \mathbf{d} = 20 \cos 30^\circ \times 4 = 20 \times \frac{\sqrt{3}}{2} \times 4 = 40\) J.
104. If a force of 15 N is applied at an angle of \(45^\circ\) to the horizontal, and the object moves horizontally a distance of 6 m, what is the work done?
ⓐ. 91.45 J
ⓑ. 61.66 J
ⓒ. 63.66 J
ⓓ. 51.57 J
Correct Answer: 63.66 J
Explanation: The work done \( W \) is calculated using the horizontal component of the force \( F \):
\[ F_{\text{horizontal}} = 15 \cdot \cos 45^\circ = 15 \cdot \frac{\sqrt{2}}{2} = 10.61 \text{ N} \]
Now, calculate the work done:
\[ W = F_{\text{horizontal}} \cdot d = 10.61 \cdot 6 = 63.66 \text{ J} \]
105. A force of 10 N is applied horizontally to move an object 5 m along a frictionless surface. What is the work done?
ⓐ. 0 J
ⓑ. 50 J
ⓒ. 10 J
ⓓ. 5 J
Correct Answer: 50 J
Explanation: Since the force and displacement are in the same direction, \(W = F \cdot d = 10 \times 5 = 50\) J.
106. A force of 15 N is applied vertically upwards on an object that moves vertically upwards a distance of 3 m. What is the work done?
ⓐ. 45 J
ⓑ. 30 J
ⓒ. 15 J
ⓓ. 0 J
Correct Answer: 0 J
Explanation: The force and displacement are perpendicular (\(\theta = 90^\circ\)), so \(W = F \cdot d \cdot \cos 90^\circ = 15 \times 3 \times 0 = 0\) J.
107. A force of 8 N is applied at an angle of \(30^\circ\) above the horizontal to move an object 4 m horizontally. What is the work done?
ⓐ. 27 J
ⓑ. 32 J
ⓒ. 16 J
ⓓ. 12 J
Correct Answer: 27 J
Explanation: The horizontal component of the force is \(8 \cos 30^\circ = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}\) N. Therefore, \(W = 4\sqrt{3} \times 4 = 16\sqrt{3} \approx 27.7\) J.
108. If a force of 20 N is applied horizontally to move an object 5 m vertically upwards, what is the work done?
ⓐ. 0 J
ⓑ. 100 J
ⓒ. 20 J
ⓓ. 5 J
Correct Answer: 0 J
Explanation: The force is applied horizontally, but the displacement is vertically upwards, so \(\theta = 90^\circ\) and \(W = F \cdot d \cdot \cos 90^\circ = 20 \times 5 \times 0 = 0\) J.
109. A force of 12 N is applied at an angle of \(60^\circ\) to the horizontal to move an object 8 m horizontally. What is the work done?
ⓐ. 48 J
ⓑ. 24 J
ⓒ. 96 J
ⓓ. 64 J
Correct Answer: 48 J
Explanation: The horizontal component of the force is \(12 \cos 60^\circ = 12 \times 0.5 = 6\) N. Therefore, \(W = 6 \times 8 = 48\) J.
110. A force of 10 N is applied at an angle of \(45^\circ\) to the horizontal to move an object 6 m horizontally. What is the work done?
ⓐ. 30 J
ⓑ. 42.4 J
ⓒ. 60 J
ⓓ. 10 J
Correct Answer: 42.4 J
Explanation: The horizontal component of the force is \(10 \cos 45^\circ = 10 \times \frac{\sqrt{2}}{2} = 5\sqrt{2}\) N. Therefore, \(W = 5\sqrt{2} \times 6 = 30\sqrt{2} \approx 42.4\) J.
111. A force of 15 N is applied vertically downwards on an object that moves vertically upwards a distance of 4 m. What is the work done?
ⓐ. -60 J
ⓑ. 60 J
ⓒ. -45 J
ⓓ. 45 J
Correct Answer: -60 J
Explanation: The force and displacement are in opposite directions, so the work done is negative: \(W = F \cdot d \cdot \cos 180^\circ = 15 \times 4 \times (-1) = -60\) J.
112. A force of 25 N is applied at an angle of \(60^\circ\) to the horizontal to move an object 10 m horizontally. What is the work done?
ⓐ. 100 J
ⓑ. 125 J
ⓒ. 150 J
ⓓ. 175 J
Correct Answer: 125 J
Explanation: The horizontal component of the force is \(25 \cos 60^\circ = 25 \times 0.5 = 12.5\) N. Therefore, \(W = 12.5 \times 10 = 125\) J.
113. If a force of 30 N is applied horizontally to move an object 8 m vertically upwards, what is the work done?
ⓐ. 0 J
ⓑ. 240 J
ⓒ. 120 J
ⓓ. 240 N
Correct Answer: 0 J
Explanation: The force is applied horizontally, but the displacement is vertically upwards, so \(\theta = 90^\circ\) and \(W = F \cdot d \cdot \cos 90^\circ = 30 \times 8 \times 0 = 0\) J.
114. A force of 18 N is applied vertically downwards on an object that moves vertically downwards a distance of 5 m. What is the work done?
ⓐ. -90 J
ⓑ. 90 J
ⓒ. -45 J
ⓓ. 45 J
Correct Answer: 90 J
Explanation: The force and displacement are in the same direction, so \(W = F \cdot d \cdot \cos 0^\circ = 18 \times 5 \times 1 = 90\) J.
115. A force of 40 N is applied at an angle of \(30^\circ\) to the horizontal to move an object 12 m horizontally. What is the work done?
ⓐ. 240 J
ⓑ. 400 J
ⓒ. 480 J
ⓓ. 600 J
Correct Answer: 240 J
Explanation: The horizontal component of the force is \(40 \cos 30^\circ = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3}\) N. Therefore, \(W = 20\sqrt{3} \times 12 = 240\) J.
116. A force of 12 N is applied horizontally to move an object 8 m horizontally. What is the work done?
ⓐ. 0 J
ⓑ. 12 J
ⓒ. 96 J
ⓓ. 20 J
Correct Answer: 96 J
Explanation: Since the force and displacement are in the same direction, \(W = F \cdot d = 12 \times 8 = 96\) J.
117. A force of 20 N is applied vertically upwards on an object that moves vertically upwards a distance of 5 m. What is the work done?
ⓐ. 0 J
ⓑ. 100 J
ⓒ. 20 J
ⓓ. 5 J
Correct Answer: 100 J
Explanation: The force and displacement are in the same direction, so \(W = F \cdot d = 20 \times 5 = 100\) J.
118. A force of 30 N is applied at an angle of \(45^\circ\) to the horizontal to move an object 10 m horizontally. What is the work done?
ⓐ. 300 J
ⓑ. 150 J
ⓒ. 210 J
ⓓ. 212.13 J
Correct Answer: 212.13 J
Explanation: The horizontal component of the force is \(30 \cos 45^\circ = 30 \times \frac{\sqrt{2}}{2} = 15\sqrt{2}\) N. Therefore, \(W = 15\sqrt{2} \times 10 = 150\sqrt{2} \approx 212.13\) J.
119. A force of 25 N is applied vertically downwards on an object that moves vertically upwards a distance of 6 m. What is the work done?
ⓐ. -150 J
ⓑ. 150 J
ⓒ. -125 J
ⓓ. 125 J
Correct Answer: -150 J
Explanation: The force and displacement are in opposite directions, so the work done is negative: \(W = F \cdot d \cdot \cos 180^\circ = 25 \times 6 \times (-1) = -150\) J.
120. If a force of 15 N is applied at an angle of \(30^\circ\) above the horizontal to move an object 4 m horizontally, what is the work done?
ⓐ. 60 J
ⓑ. 30 J
ⓒ. 20 J
ⓓ. 10 J
Correct Answer: 30 J
Explanation: The horizontal component of the force is \(15 \cos 30^\circ = 15 \times \frac{\sqrt{3}}{2} = 7.5\sqrt{3}\) N. Therefore, \(W = 7.5\sqrt{3} \times 4 = 30\) J.
121. A force of 18 N is applied horizontally to move an object 6 m vertically upwards. What is the work done?
ⓐ. 0 J
ⓑ. 18 J
ⓒ. 108 J
ⓓ. 6 J
Correct Answer: 0 J
Explanation: The force is applied horizontally, but the displacement is vertically upwards, so \(\theta = 90^\circ\) and \(W = F \cdot d \cdot \cos 90^\circ = 18 \times 6 \times 0 = 0\) J.
122. A force of 40 N is applied at an angle of \(60^\circ\) to the horizontal to move an object 5 m vertically upwards. What is the work done?
ⓐ. 0 J
ⓑ. 200 J
ⓒ. 100 J
ⓓ. 20 J
Correct Answer: 0 J
Explanation: The force is applied at an angle, but the displacement is vertically upwards, so the horizontal component of the force does no work on the object, \(W = F \cdot d \cdot \cos 60^\circ = 40 \times 5 \times 0 = 0\) J.
123. Define work done by a variable force.
ⓐ. Work done by a force that changes direction
ⓑ. Work done by a force that varies in magnitude
ⓒ. Work done by a force that moves in a circular path
ⓓ. Work done by a force that remains constant
Correct Answer: Work done by a force that varies in magnitude
Explanation: Work done by a variable force is defined as the integral of the force over the displacement: \(W = \int \mathbf{F}(x) \cdot d\mathbf{x}\).
124. A force varies with displacement as \(F(x) = 2x + 3\), where \(x\) is in meters. Calculate the work done by this force over a displacement from \(x = 1\) m to \(x = 4\) m.
ⓐ. 15 J
ⓑ. 20 J
ⓒ. 25 J
ⓓ. 24 J
Correct Answer: 24 J
Explanation: \(W = \int_{1}^{4} (2x + 3) \, dx = \left[ x^2 + 3x \right]_{1}^{4} = (16 + 12) – (1 + 3) = 28 – 4 = 24\) J.
125. What is the SI unit of work done by a variable force?
ⓐ. N
ⓑ. J
ⓒ. W
ⓓ. kg m/s²
Correct Answer: J
Explanation: The SI unit of work is joule (J), which is defined as 1 newton-meter (N·m).
126. If a force \(F(x) = 3x^2\) acts on an object displacing it from \(x = 0\) to \(x = 2\) meters, what is the work done?
ⓐ. 8 J
ⓑ. 12 J
ⓒ. 16 J
ⓓ. 24 J
Correct Answer: 8 J
Explanation: \(W = \int_{0}^{2} 3x^2 \, dx = \left[ x^3 \right]_{0}^{2} = 8\) J.
127. A force varies with position as \(F(x) = 4 + \frac{2}{x}\), where \(x\) is in meters. Calculate the work done by this force over a displacement from \(x = 1\) m to \(x = 3\) m.
ⓐ. 2 J
ⓑ. 3 J
ⓒ. 4 J
ⓓ. 5 J
Correct Answer: 3 J
Explanation: \(W = \int_{1}^{3} \left( 4 + \frac{2}{x} \right) \, dx = \left[ 4x + 2 \ln x \right]_{1}^{3} = (12 + 2 \ln 3) – (4 + 2 \ln 1) = 12 + 2 \ln 3 – 4 = 8 + 2 \ln 3\) J.
128. Define the work-energy theorem.
ⓐ. Work done equals the change in potential energy
ⓑ. Work done equals the change in kinetic energy
ⓒ. Work done equals the change in gravitational energy
ⓓ. Work done equals the change in thermal energy
Correct Answer: Work done equals the change in kinetic energy
Explanation: According to the work-energy theorem, the work done by the net force on an object equals the change in its kinetic energy.
129. If a force varies with displacement as \(F(x) = 6x\), where \(x\) is in meters, and it moves from \(x = 1\) m to \(x = 4\) m, what is the work done?
ⓐ. 48 J
ⓑ. 45 J
ⓒ. 60 J
ⓓ. 72 J
Correct Answer: 45 J
Explanation: \(W = \int_{1}^{4} 6x \, dx = \left[ 3x^2 \right]_{1}^{4} = 48 – 3 = 45\) J.
130. Calculate the work done by a force \(F(x) = 2x\) over a displacement from \(x = 0\) to \(x = 5\) meters.
ⓐ. 25 J
ⓑ. 50 J
ⓒ. 75 J
ⓓ. 100 J
Correct Answer: 25 J
Explanation: \(W = \int_{0}^{5} 2x \, dx = \left[ x^2 \right]_{0}^{5} = 25\) J.
131. A force \(F(x) = \frac{6}{x}\) acts on an object displacing it from \(x = 1\) m to \(x = 3\) m. Calculate the work done.
ⓐ. 8 J
ⓑ. 9 J
ⓒ. 10 J
ⓓ. 12 J
Correct Answer: 9 J
Explanation: \(W = \int_{1}^{3} \frac{6}{x} \, dx = \left[ 6 \ln x \right]_{1}^{3} = 6 \ln 3 – 6 \ln 1 = 6 \ln 3\) J.
132. If a force \(F(x) = 4x^3\) acts on an object displacing it from \(x = 0\) to \(x = 2\) meters, what is the work done?
ⓐ. 8 J
ⓑ. 16 J
ⓒ. 32 J
ⓓ. 64 J
Correct Answer: 16 J
Explanation: \(W = \int_{0}^{2} 4x^3 \, dx = \left[ x^4 \right]_{0}^{2} = 16\) J.
133. Calculate the work done by a force \(F(x) = 5\) over a displacement from \(x = 1\) m to \(x = 4\) m.
ⓐ. 15 J
ⓑ. 20 J
ⓒ. 25 J
ⓓ. 30 J
Correct Answer: 20 J
Explanation: \(W = \int_{1}^{4} 5 \, dx = \left[ 5x \right]_{1}^{4} = 20\) J.
134. A force \(F(x) = 3x^2 + 2x\) acts on an object displacing it from \(x = 1\) m to \(x = 3\) m. Calculate the work done.
ⓐ. 23 J
ⓑ. 28 J
ⓒ. 31 J
ⓓ. 35 J
Correct Answer: 35 J
Explanation: \(W = \int_{1}^{3} (3x^2 + 2x) \, dx = \left[ x^3 + x^2 \right]_{1}^{3} = (27 + 9) – (1 + 1) = 35\) J.
135. Calculate the work done by a force \(F(x) = \frac{10}{x}\) over a displacement from \(x = 2\) m to \(x = 5\) m.
ⓐ. 8 J
ⓑ. 12 J
ⓒ. 16 J
ⓓ. 20 J
Correct Answer: 8 J
Explanation: \(W = \int_{2}^{5} \frac{10}{x} \, dx = \left[ 10 \ln x \right]_{2}^{5} = 10 \ln 5 – 10 \ln 2\) J.
136. If a force \(F(x) = 2x^2 + 3x\) acts on an object displacing it from \(x = 0\) to \(x = 4\) meters, what is the work done?
ⓐ. 24 J
ⓑ. 32 J
ⓒ. 48 J
ⓓ. 64 J
Correct Answer: 48 J
Explanation: \(W = \int_{0}^{4} (2x^2 + 3x) \, dx = \left[ \frac{2x^3}{3} + \frac{3x^2}{2} \right]_{0}^{4} = \left( \frac{2 \cdot 64}{3} + \frac{3 \cdot 16}{2} \right) – (0) = \frac{128}{3} + 24 = \frac{128 + 72}{3} = \frac{200}{3} = 48.\overline{3}\) J (approximately 48 J).
137. A force \(F(x) = 4x\) acts on an object displacing it from \(x = 1\) m to \(x = 3\) m. Calculate the work done.
ⓐ. 16 J
ⓑ. 24 J
ⓒ. 32 J
ⓓ. 40 J
Correct Answer: 16 J
Explanation: \(W = \int_{1}^{3} 4x \, dx = \left[ 2x^2 \right]_{1}^{3} = 18 – 2 = 16\) J.
138. Calculate the work done by a force \(F(x) = 3\) over a displacement from \(x = 0\) m to \(x = 5\) m.
ⓐ. 15 J
ⓑ. 18 J
ⓒ. 20 J
ⓓ. 25 J
Correct Answer: 15 J
Explanation: \(W = \int_{0}^{5} 3 \, dx = \left[ 3x \right]_{0}^{5} = 15\) J.
139. A force \(F(x) = \frac{8}{x}\) acts on an object displacing it from \(x = 2\) m to \(x = 4\) m. Calculate the work done.
ⓐ. 6 J
ⓑ. 8 J
ⓒ. 10 J
ⓓ. 12 J
Correct Answer: 8 J
Explanation: \(W = \int_{2}^{4} \frac{8}{x} \, dx = \left[ 8 \ln x \right]_{2}^{4} = 8 \ln 4 – 8 \ln 2 = 8 (\ln 4 – \ln 2) = 8 \ln 2\) J.
140. What is kinetic energy?
ⓐ. Energy stored in an object due to its position
ⓑ. Energy stored in an object due to its motion
ⓒ. Energy stored in an object due to its temperature
ⓓ. Energy stored in an object due to its shape
Correct Answer: Energy stored in an object due to its motion
Explanation: Kinetic energy is the energy possessed by an object due to its motion.
141. What is the formula for kinetic energy (KE) of an object of mass \(m\) moving with velocity \(v\)?
ⓐ. \( KE = \frac{1}{2} mv^2 \)
ⓑ. \( KE = \frac{1}{2} mv \)
ⓒ. \( KE = \frac{1}{2} m + v \)
ⓓ. \( KE = mv^2 \)
Correct Answer: \( KE = \frac{1}{2} mv^2 \)
Explanation: Kinetic energy is calculated using the formula \( KE = \frac{1}{2} mv^2 \), where \(m\) is the mass of the object and \(v\) is its velocity.
142. If an object has twice the velocity, how does its kinetic energy change, assuming its mass remains constant?
ⓐ. It increases by a factor of 4
ⓑ. It decreases by a factor of 2
ⓒ. It increases by a factor of 2
ⓓ. It remains the same
Correct Answer: It increases by a factor of 4
Explanation: Kinetic energy is directly proportional to the square of velocity, so if velocity doubles, kinetic energy increases by a factor of \(2^2 = 4\).
143. Which SI unit is used to measure kinetic energy?
ⓐ. Newton (N)
ⓑ. Joule (J)
ⓒ. Watt (W)
ⓓ. Meter (m)
Correct Answer: Joule (J)
Explanation: The SI unit of kinetic energy (as well as all forms of energy) is the joule (J).
144. An object of mass 2 kg is moving with a velocity of 3 m/s. What is its kinetic energy?
ⓐ. 3 J
ⓑ. 6 J
ⓒ. 9 J
ⓓ. 18 J
Correct Answer: 9 J
Explanation: \( KE = \frac{1}{2} mv^2 = \frac{1}{2} \times 2 \times 3^2 = \frac{1}{2} \times 2 \times 9 = 9 \) J.
145. Kinetic energy depends on which of the following factors?
ⓐ. Shape of the object
ⓑ. Volume of the object
ⓒ. Mass and speed of the object
ⓓ. Temperature of the object
Correct Answer: Mass and speed of the object
Explanation: Kinetic energy depends on both the mass (m) and the square of the speed (v) of the object.
146. If the velocity of an object is halved, how does its kinetic energy change, assuming its mass remains constant?
ⓐ. It decreases by a factor of 4
ⓑ. It decreases by a factor of 2
ⓒ. It decreases by a factor of \( \frac{1}{4} \)
ⓓ. It decreases by a factor of \( \frac{1}{2} \)
Correct Answer: It decreases by a factor of \( \frac{1}{4} \)
Explanation: Kinetic energy is proportional to the square of velocity, so if velocity is halved, kinetic energy decreases by a factor of \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \).
147. Which term describes the energy associated with the motion of an object?
ⓐ. Potential energy
ⓑ. Kinetic energy
ⓒ. Thermal energy
ⓓ. Chemical energy
Correct Answer: Kinetic energy
Explanation: Kinetic energy is the energy associated with the motion of an object.
148. An object of mass 5 kg has a kinetic energy of 100 J. What is its velocity?
ⓐ. 2 m/s
ⓑ. 4 m/s
ⓒ. 6 m/s
ⓓ. 8 m/s
Correct Answer: 4 m/s
Explanation: \( KE = \frac{1}{2} mv^2 \). Solving for \(v\), \( v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 100}{5}} = \sqrt{40} = 2 \sqrt{10} \approx 4 \) m/s.
149. Which of the following is true about kinetic energy?
ⓐ. It depends only on the mass of the object
ⓑ. It depends only on the speed of the object
ⓒ. It depends on both mass and speed of the object
ⓓ. It depends on the shape of the object
Correct Answer: It depends on both mass and speed of the object
Explanation: Kinetic energy depends on both the mass and the square of the speed of the object.
150. If an object’s velocity triples, how does its kinetic energy change, assuming its mass remains constant?
ⓐ. It increases by a factor of 3
ⓑ. It increases by a factor of 6
ⓒ. It increases by a factor of 9
ⓓ. It increases by a factor of 27
Correct Answer: It increases by a factor of 9
Explanation: Kinetic energy is proportional to the square of velocity, so if velocity triples, kinetic energy increases by a factor of \(3^2 = 9\).
151. What is the relationship between work \(W\) done on an object and its change in kinetic energy \( \Delta KE \)?
ⓐ. \( W = \Delta KE \)
ⓑ. \( W = \frac{1}{2} \Delta KE \)
ⓒ. \( W = 2 \Delta KE \)
ⓓ. \( W = \frac{1}{2} KE \)
Correct Answer: \( W = \Delta KE \)
Explanation: According to the work-energy theorem, the work done on an object is equal to its change in kinetic energy.
152. An object starts from rest and reaches a velocity of 10 m/s due to a constant force. How much work was done on the object if its mass is 2 kg?
ⓐ. 100 J
ⓑ. 50 J
ⓒ. 20 J
ⓓ. 10 J
Correct Answer: 100 J
Explanation: \( W = \frac{1}{2} mv^2 = \frac{1}{2} \times 2 \times (10)^2 = 100 \) J.
153. If the work done on an object is negative, what does this imply about its kinetic energy change?
ⓐ. Kinetic energy decreases
ⓑ. Kinetic energy increases
ⓒ. Kinetic energy remains constant
ⓓ. Kinetic energy becomes zero
Correct Answer: Kinetic energy decreases
Explanation: Negative work means the force applied is opposite to the direction of motion, resulting in a decrease in kinetic energy.
154. A force of 5 N acts on an object, moving it a distance of 10 m. If the object’s initial speed was 2 m/s and its mass is 3 kg, what is its final kinetic energy?
ⓐ. 45 J
ⓑ. 56 J
ⓒ. 65 J
ⓓ. 90 J
Correct Answer: 56 J
Explanation: First, calculate the work done: \( W = F \cdot d = 5 \cdot 10 = 50 \) J. Then, calculate the initial kinetic energy: \( KE_{\text{initial}} = \frac{1}{2} mv^2 = \frac{1}{2} \times 3 \times (2)^2 = 6 \) J. Finally, the final kinetic energy \( KE_{\text{final}} = KE_{\text{initial}} + W = 6 + 50 = 56 \) J.
155. If a force of 10 N acts on an object to move it 5 m horizontally, what is the work done on the object? Assume the force is horizontal and the object moves horizontally.
ⓐ. 5 J
ⓑ. 10 J
ⓒ. 50 J
ⓓ. 80 J
Correct Answer: 50 J
Explanation: \( W = F \cdot d \cos \theta = 10 \cdot 5 \cdot \cos(0^\circ) = 50 \) J, where \( \theta \) is the angle between the force and the displacement (0 degrees for horizontal motion).
156. An object of mass 4 kg is moving with a velocity of 6 m/s. How much work is required to bring it to rest?
ⓐ. 72 J
ⓑ. 108 J
ⓒ. 144 J
ⓓ. 216 J
Correct Answer: 72 J
Explanation: Work required to bring the object to rest \( W = \frac{1}{2} mv^2 = \frac{1}{2} \times 4 \times (6)^2 = 72 \) J.
157. If a 1000 kg car accelerates from 0 to 20 m/s in 10 seconds, how much work was done on it?
ⓐ. 20,000 J
ⓑ. 40,000 J
ⓒ. 60,000 J
ⓓ. 80,000 J
Correct Answer: 20,000 J
Explanation: \( W = \frac{1}{2} mv^2 = \frac{1}{2} \times 1000 \times (20)^2 = 20,000 \) J.
158. A force of 15 N acts on an object, causing it to accelerate at 5 m/s². If the object moves a distance of 10 m, how much work is done?
ⓐ. 100 J
ⓑ. 120 J
ⓒ. 150 J
ⓓ. 525 J
Correct Answer: 150 J
Explanation: First, calculate the final velocity using \( v^2 = u^2 + 2as \), then use \( W = F \cdot s \).
159. An object of mass 5 kg is moving with a velocity of 10 m/s. How much work is done to double its kinetic energy?
ⓐ. 250 J
ⓑ. 500 J
ⓒ. 750 J
ⓓ. 1000 J
Correct Answer: 500 J
Explanation: To double the kinetic energy, \( KE_{\text{final}} = 2 \times KE_{\text{initial}} = 2 \times 250 = 500 \) J.
160. In a car crash, which type of energy is primarily responsible for causing damage?
ⓐ. Potential energy
ⓑ. Kinetic energy
ⓒ. Thermal energy
ⓓ. Chemical energy
Correct Answer: Kinetic energy
Explanation: During a car crash, the kinetic energy of the moving vehicle is primarily responsible for causing damage upon impact.
161. A bullet fired from a gun possesses kinetic energy due to its:
ⓐ. Position
ⓑ. Mass
ⓒ. Temperature
ⓓ. Motion
Correct Answer: Motion
Explanation: The kinetic energy of a bullet fired from a gun is due to its motion.
162. Which type of energy do wind turbines convert into electrical energy?
ⓐ. Kinetic energy
ⓑ. Potential energy
ⓒ. Thermal energy
ⓓ. Chemical energy
Correct Answer: Kinetic energy
Explanation: Wind turbines convert the kinetic energy of moving air (wind) into electrical energy.
163. When a hammer strikes a nail, the work done is primarily converted into:
ⓐ. Kinetic energy
ⓑ. Potential energy
ⓒ. Sound energy
ⓓ. Thermal energy
Correct Answer: Thermal energy
Explanation: When a hammer strikes a nail, the work done is primarily converted into thermal energy due to friction between the hammer and the nail.
164. Which of the following is an example of kinetic energy being converted into potential energy?
ⓐ. A pendulum at its highest point
ⓑ. A car moving at a constant speed
ⓒ. A spinning top
ⓓ. A ball rolling down a slope
Correct Answer: A pendulum at its highest point
Explanation: In a pendulum, kinetic energy is converted into potential energy when it reaches its highest point.
165. In a roller coaster ride, which form of energy changes most significantly throughout the ride?
ⓐ. Chemical energy
ⓑ. Kinetic energy
ⓒ. Potential energy
ⓓ. Electrical energy
Correct Answer: Potential energy
Explanation: In a roller coaster ride, potential energy changes significantly as the coaster moves between high and low points.
166. Which energy transformation occurs when a person jumps from a diving board into a swimming pool?
ⓐ. Potential energy to kinetic energy
ⓑ. Kinetic energy to potential energy
ⓒ. Thermal energy to mechanical energy
ⓓ. Electrical energy to sound energy
Correct Answer: Potential energy to kinetic energy
Explanation: As a person jumps from a diving board, potential energy due to height is converted into kinetic energy during the descent.
167. When a tennis ball is hit by a racket, which energy transformation primarily occurs?
ⓐ. Thermal energy to kinetic energy
ⓑ. Electrical energy to potential energy
ⓒ. Chemical energy to mechanical energy
ⓓ. Potential energy to kinetic energy
Correct Answer: Potential energy to kinetic energy
Explanation: When a tennis ball is hit by a racket, potential energy stored in the stretched strings of the racket is converted into kinetic energy of the ball.
168. Which device uses the conversion of kinetic energy into electrical energy?
ⓐ. Solar panel
ⓑ. Wind turbine
ⓒ. Battery
ⓓ. Light bulb
Correct Answer: Wind turbine
Explanation: Wind turbines use the kinetic energy of wind to generate electrical energy.
169. In a hydroelectric power plant, which form of energy is used to generate electricity?
ⓐ. Chemical energy
ⓑ. Kinetic energy
ⓒ. Potential energy
ⓓ. Nuclear energy
Correct Answer: Potential energy
Explanation: In a hydroelectric power plant, potential energy of water stored in a reservoir is converted into kinetic energy as it falls through turbines, generating electrical energy.
170. When a bicycle is pedaled, which form of energy is primarily responsible for propelling the bicycle forward?
ⓐ. Chemical energy
ⓑ. Electrical energy
ⓒ. Kinetic energy
ⓓ. Potential energy
Correct Answer: Kinetic energy
Explanation: Pedaling a bicycle converts chemical energy from the rider’s muscles into kinetic energy, propelling the bicycle forward.
171. What does the Work-Energy Theorem state?
ⓐ. The work done on an object is equal to its displacement times its acceleration.
ⓑ. The work done on an object is equal to the change in its kinetic energy.
ⓒ. The work done on an object is equal to its gravitational potential energy.
ⓓ. The work done on an object is equal to its momentum.
Correct Answer: The work done on an object is equal to the change in its kinetic energy.
Explanation: According to the Work-Energy Theorem, the work done on an object by the net force is equal to the change in its kinetic energy.
172. If a force of 20 N acts on an object and moves it a distance of 5 m, how much work is done according to the Work-Energy Theorem?
ⓐ. 100 J
ⓑ. 80 J
ⓒ. 60 J
ⓓ. 40 J
Correct Answer: 100 J
Explanation: Work done \( W = F \cdot d \cos \theta = 20 \times 5 \times \cos(0^\circ) = 100 \) J, where \( \theta \) is the angle between the force and displacement (0 degrees for horizontal motion).
173. Which of the following best describes the Work-Energy Theorem?
ⓐ. Work done on an object increases its speed.
ⓑ. Work done on an object is equal to the force applied to it.
ⓒ. Work done on an object changes its shape.
ⓓ. Work done on an object changes its kinetic energy.
Correct Answer: Work done on an object changes its kinetic energy.
Explanation: The Work-Energy Theorem states that the work done on an object is equal to the change in its kinetic energy.
174. If the net work done on an object is zero, what can be said about its kinetic energy?
ⓐ. Kinetic energy increases.
ⓑ. Kinetic energy decreases.
ⓒ. Kinetic energy remains constant.
ⓓ. Kinetic energy becomes zero.
Correct Answer: Kinetic energy remains constant.
Explanation: If the net work done on an object is zero, its kinetic energy remains constant according to the Work-Energy Theorem.
175. A car accelerates from rest to 30 m/s. If the work done on the car by the engine is 150,000 J, what is its mass?
ⓐ. 5000 kg
ⓑ. 2500 kg
ⓒ. 7500 kg
ⓓ. 10000 kg
Correct Answer: 2500 kg
Explanation: Work done \( W = \frac{1}{2} mv^2 \), solve for mass.
176. When a force acts on an object, causing it to move in the direction of the force, what happens to the object’s kinetic energy according to the Work-Energy Theorem?
ⓐ. Increases
ⓑ. Decreases
ⓒ. Remains constant
ⓓ. Becomes zero
Correct Answer: Increases
Explanation: If a force acts on an object causing it to move in the direction of the force, the object’s kinetic energy increases as per the Work-Energy Theorem.
177. If a ball is thrown vertically upwards and reaches its maximum height, what can be said about its kinetic energy according to the Work-Energy Theorem?
ⓐ. It increases
ⓑ. It decreases
ⓒ. It remains constant
ⓓ. It becomes zero
Correct Answer: It becomes zero
Explanation: At the highest point, the ball’s velocity is zero, hence its kinetic energy is zero.
178. A spring is compressed by a force. When the spring is released, what type of energy does it primarily convert into, according to the Work-Energy Theorem?
ⓐ. Kinetic energy
ⓑ. Potential energy
ⓒ. Electrical energy
ⓓ. Chemical energy
Correct Answer: Kinetic energy
Explanation: A compressed spring primarily converts potential energy into kinetic energy when released.
179. If the work done on an object is negative according to the Work-Energy Theorem, what does this imply about its kinetic energy change?
ⓐ. Kinetic energy decreases
ⓑ. Kinetic energy increases
ⓒ. Kinetic energy remains constant
ⓓ. Kinetic energy becomes zero
Correct Answer: Kinetic energy decreases
Explanation: Negative work implies that the force applied is opposite to the direction of motion, resulting in a decrease in kinetic energy.
180. Which theorem is used to relate the work done on an object to its kinetic energy change?
ⓐ. Newton’s First Theorem
ⓑ. Work-Power Theorem
ⓒ. Work-Energy Theorem
ⓓ. Hooke’s Theorem
Correct Answer: Work-Energy Theorem
Explanation: The Work-Energy Theorem relates the work done on an object by the net force to its change in kinetic energy.
181. State the formula derived from the Work-Energy Theorem that relates the work done by a constant force to the change in kinetic energy.
ⓐ. \( W = F \cdot d \)
ⓑ. \( W = \frac{1}{2} m v^2 \)
ⓒ. \( W = \Delta KE \)
ⓓ. \( W = \Delta PE \)
Correct Answer: \( W = \Delta KE \)
Explanation: According to the Work-Energy Theorem, the work done by the net force on an object is equal to the change in its kinetic energy.
182. A 2 kg object initially at rest is acted upon by a constant force of 10 N for a distance of 5 m. Calculate the change in kinetic energy using the Work-Energy Theorem.
ⓐ. 25 J
ⓑ. 50 J
ⓒ. 100 J
ⓓ. 200 J
Correct Answer: 50 J
Explanation: Work done \( W = F \cdot d = 10 \times 5 = 50 \) J. Change in kinetic energy \( \Delta KE = W = 50 \) J.
183. What principle does the Work-Energy Theorem illustrate about energy in mechanical systems?
ⓐ. Conservation of energy
ⓑ. First law of thermodynamics
ⓒ. Second law of thermodynamics
ⓓ. Energy dissipation
Correct Answer: Conservation of energy
Explanation: The Work-Energy Theorem illustrates the principle of conservation of energy in mechanical systems.
184. Which concept is fundamental to the application of the Work-Energy Theorem to various mechanical situations?
ⓐ. Newton’s laws of motion
ⓑ. Hooke’s law
ⓒ. Conservation of momentum
ⓓ. Law of universal gravitation
Correct Answer: Newton’s laws of motion
Explanation: The application of the Work-Energy Theorem relies on Newton’s laws of motion to relate forces and motion to changes in kinetic energy.
185. In the context of the Work-Energy Theorem, what happens if the net work done on an object is zero?
ⓐ. Kinetic energy decreases
ⓑ. Kinetic energy increases
ⓒ. Kinetic energy remains constant
ⓓ. Kinetic energy becomes zero
Correct Answer: Kinetic energy remains constant
Explanation: If the net work done on an object is zero according to the Work-Energy Theorem, its kinetic energy remains constant.
186. When using the Work-Energy Theorem, why is it important to consider all forces acting on an object?
ⓐ. To calculate the object’s velocity
ⓑ. To ensure conservation of energy
ⓒ. To determine its potential energy
ⓓ. To find the force of gravity
Correct Answer: To ensure conservation of energy
Explanation: Considering all forces ensures that the total work done on the object accounts for changes in its kinetic energy and potential energy, maintaining conservation of energy.
187. If a force does negative work on an object, what does this imply according to the Work-Energy Theorem?
ⓐ. The object’s kinetic energy decreases
ⓑ. The object’s kinetic energy increases
ⓒ. The object’s potential energy decreases
ⓓ. The object’s potential energy increases
Correct Answer: The object’s kinetic energy decreases
Explanation: Negative work done by a force implies that the force opposes the direction of motion, resulting in a decrease in kinetic energy according to the Work-Energy Theorem.
188. Which theorem relates the work done on an object to its change in kinetic energy?
ⓐ. Work-Power Theorem
ⓑ. Work-Energy Theorem
ⓒ. Conservation of Energy Theorem
ⓓ. Newton’s Second Law
Correct Answer: Work-Energy Theorem
Explanation: The Work-Energy Theorem specifically relates the work done on an object by the net force to its change in kinetic energy.
189. How does the Work-Energy Theorem contribute to solving problems involving motion and forces?
ⓐ. By calculating potential energy
ⓑ. By determining acceleration
ⓒ. By relating work to changes in energy
ⓓ. By analyzing frictional forces
Correct Answer: By relating work to changes in energy
Explanation: The Work-Energy Theorem provides a direct way to relate the work done on an object to changes in its kinetic energy, aiding in the analysis of motion and forces.
190. In the context of the Work-Energy Theorem, what does the term “mechanical energy” refer to?
ⓐ. Sum of kinetic and potential energies
ⓑ. Sum of kinetic and thermal energies
ⓒ. Sum of potential and thermal energies
ⓓ. Sum of kinetic and chemical energies
Correct Answer: Sum of kinetic and potential energies
Explanation: Mechanical energy, in the context of the Work-Energy Theorem, refers to the sum of an object’s kinetic and potential energies.
191. What principle states that in the absence of non-conservative forces, the total mechanical energy of a system remains constant?
ⓐ. Conservation of momentum
ⓑ. Newton’s third law
ⓒ. Conservation of mechanical energy
ⓓ. Second law of thermodynamics
Correct Answer: Conservation of mechanical energy
Explanation: Conservation of mechanical energy states that in a system without non-conservative forces (like friction), the total mechanical energy (sum of kinetic and potential energies) remains constant.
192. A pendulum swings back and forth. According to the conservation of mechanical energy, what happens to its total mechanical energy over time?
ⓐ. It increases
ⓑ. It decreases
ⓒ. It remains constant
ⓓ. It becomes zero
Correct Answer: It remains constant
Explanation: In the absence of friction or other non-conservative forces, a pendulum’s total mechanical energy (kinetic + potential) remains constant due to conservation of mechanical energy.
193. If a ball rolls down a frictionless hill, how does its kinetic energy change compared to its potential energy according to the conservation of mechanical energy?
ⓐ. Kinetic energy increases, potential energy decreases
ⓑ. Kinetic energy decreases, potential energy increases
ⓒ. Kinetic energy remains constant, potential energy decreases
ⓓ. Kinetic energy decreases, potential energy remains constant
Correct Answer: Kinetic energy increases, potential energy decreases
Explanation: As the ball rolls down the hill, potential energy (due to height) decreases and kinetic energy (due to motion) increases, while the total mechanical energy remains constant.
194. Which type of energy is considered when applying the conservation of mechanical energy?
ⓐ. Electrical energy
ⓑ. Thermal energy
ⓒ. Kinetic and potential energy
ⓓ. Magnetic energy
Correct Answer: Kinetic and potential energy
Explanation: Conservation of mechanical energy specifically deals with the sum of kinetic and potential energies in a system.
195. A roller coaster car starts from rest at the top of a hill. As it descends, which statement best describes the conservation of mechanical energy?
ⓐ. Mechanical energy decreases
ⓑ. Mechanical energy increases
ⓒ. Mechanical energy remains constant
ⓓ. Mechanical energy becomes zero
Correct Answer: Mechanical energy remains constant
Explanation: In an idealized scenario without friction or air resistance, a roller coaster car’s total mechanical energy (kinetic + potential) remains constant as it moves from a higher to a lower position.
196. In the context of conservation of mechanical energy, what is the role of non-conservative forces like friction?
ⓐ. They increase mechanical energy
ⓑ. They decrease mechanical energy
ⓒ. They have no effect on mechanical energy
ⓓ. They convert mechanical energy into other forms
Correct Answer: They convert mechanical energy into other forms
Explanation: Non-conservative forces like friction convert mechanical energy (kinetic and potential) into other forms such as thermal energy, reducing the total mechanical energy in the system.
197. If a ball is thrown vertically upwards and reaches its highest point, what happens to its kinetic energy according to the conservation of mechanical energy?
ⓐ. It increases
ⓑ. It decreases
ⓒ. It remains constant
ⓓ. It becomes zero
Correct Answer: It decreases
Explanation: At the highest point, the ball momentarily stops moving upwards and its kinetic energy is zero. Potential energy is at its maximum due to height.
198. Which principle supports the idea that mechanical energy can neither be created nor destroyed, only transformed between kinetic and potential forms?
ⓐ. Law of conservation of energy
ⓑ. Newton’s second law
ⓒ. Law of inertia
ⓓ. Archimedes’ principle
Correct Answer: Law of conservation of energy
Explanation: The law of conservation of energy states that energy cannot be created or destroyed, only transformed. Conservation of mechanical energy is a specific application of this principle.
199. When a block slides down a frictionless inclined plane, what happens to its potential energy according to the conservation of mechanical energy?
ⓐ. It increases
ⓑ. It decreases
ⓒ. It remains constant
ⓓ. It becomes zero
Correct Answer: It decreases
Explanation: As the block slides down, potential energy (due to height) decreases while kinetic energy (due to motion) increases, keeping the total mechanical energy constant.
200. Why is the conservation of mechanical energy useful in analyzing physical systems?
ⓐ. It helps calculate electrical energy
ⓑ. It simplifies calculations by focusing on kinetic and potential energies
ⓒ. It predicts changes in magnetic fields
ⓓ. It measures gravitational forces
Correct Answer: It simplifies calculations by focusing on kinetic and potential energies
Explanation: Conservation of mechanical energy simplifies the analysis of systems by focusing on the interplay between kinetic and potential energies, especially in scenarios without non-conservative forces.
The chapter Work, Energy, and Power in Class 11 Physics (NCERT/CBSE syllabus) is highly important for scoring well in both board exams and competitive exams like JEE, NEET, and state-level entrance tests. This section of the series focuses on advanced concepts of work, energy conservation, and applications of power. In total, this chapter offers 420 MCQs with solutions, divided into 5 organized parts for systematic practice. Here in Part 2, you will find another 100 MCQs with detailed explanations to sharpen your concepts and problem-solving skills.
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- 👉 This page contains: Second set of 100 solved MCQs.
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