101. A statement says, "Energy is a vector because moving objects have direction." The best correction is that energy is
ⓐ. a vector whenever velocity is a vector
ⓑ. a scalar although velocity is a vector
ⓒ. a vector only for kinetic energy
ⓓ. a scalar only when the object is at rest
Correct Answer: a scalar although velocity is a vector
Explanation: Energy is treated as a scalar quantity in elementary mechanics. A moving object has velocity, and velocity is a vector, but the kinetic energy depends on speed squared rather than on velocity direction as a vector. Two bodies moving with the same speed in opposite directions can have the same kinetic energy. The direction of motion is important for momentum and velocity, but not for assigning a spatial direction to energy. Energy may change sign in some reference-dependent potential energy choices, but it is not a vector.
102. A body has \(K=30\,\text{J}\) and \(U=45\,\text{J}\) at a certain instant. Its mechanical energy is
ⓐ. \(15\,\text{J}\)
ⓑ. \(45\,\text{J}\)
ⓒ. \(75\,\text{J}\)
ⓓ. \(1350\,\text{J}\)
Correct Answer: \(75\,\text{J}\)
Explanation: \( \textbf{Given:} \) Kinetic energy \(K=30\,\text{J}\).
\( \textbf{Given:} \) Potential energy \(U=45\,\text{J}\).
Mechanical energy is the sum of kinetic and potential energies.
\[
E=K+U
\]
Substitute the values:
\[
E=30+45
\]
\[
E=75\,\text{J}
\]
The value is not obtained by multiplying \(K\) and \(U\); mechanical energy is an additive energy measure.
\( \textbf{Final answer:} \) The mechanical energy is \(75\,\text{J}\).
103. A smooth pendulum bob is highest at the end of its swing and lowest at the middle of its path. Ignoring air resistance, the energy form that is greatest at the lowest point is
ⓐ. kinetic energy
ⓑ. gravitational potential energy
ⓒ. elastic potential energy
ⓓ. chemical energy
Correct Answer: kinetic energy
Explanation: At the highest point of the swing, the bob has maximum height and therefore more gravitational potential energy. As it moves downward, gravity does positive work and the bob gains speed. At the lowest point, its speed is greatest, so its kinetic energy is greatest. The bob is not attached to a spring in this situation, so elastic potential energy is not the relevant form. The height-speed exchange is a typical transformation between gravitational potential energy and kinetic energy.
104. Kinetic energy of a body of mass \(m\) moving with speed \(v\) is given by
ⓐ. \(K=mv\)
ⓑ. \(K=\frac{1}{2}m^2v\)
ⓒ. \(K=\frac{v^2}{2m}\)
ⓓ. \(K=\frac{1}{2}mv^2\)
Correct Answer: \(K=\frac{1}{2}mv^2\)
Explanation: Kinetic energy is the energy associated with motion. For a body of mass \(m\) moving with speed \(v\), the expression is \(K=\frac{1}{2}mv^2\). The square of speed is important because doubling speed makes kinetic energy four times, not two times. The expression \(mv\) represents momentum magnitude only when mass and speed are used, not kinetic energy. The factor \(\frac{1}{2}\) and the square on \(v\) are both essential parts of the kinetic energy formula.
105. A \(3\,\text{kg}\) object moves with speed \(4\,\text{m s}^{-1}\). Its kinetic energy is
ⓐ. \(12\,\text{J}\)
ⓑ. \(18\,\text{J}\)
ⓒ. \(24\,\text{J}\)
ⓓ. \(48\,\text{J}\)
Correct Answer: \(24\,\text{J}\)
Explanation: \( \textbf{Given:} \) \(m=3\,\text{kg}\) and \(v=4\,\text{m s}^{-1}\).
\( \textbf{Required:} \) Kinetic energy \(K\).
\( \textbf{Formula:} \)
\[
K=\frac{1}{2}mv^2
\]
This formula applies because kinetic energy depends on mass and speed.
\( \textbf{Substitution:} \)
\[
K=\frac{1}{2}(3)(4)^2
\]
\( \textbf{Speed squared:} \)
\[
4^2=16
\]
\( \textbf{Calculation:} \)
\[
K=\frac{1}{2}(3)(16)=24\,\text{J}
\]
Using \(mv\) would give \(12\), but that is not kinetic energy.
\( \textbf{Final answer:} \) The kinetic energy is \(24\,\text{J}\).
106. If the speed of a body is doubled while its mass remains unchanged, its kinetic energy becomes
ⓐ. two times the original value
ⓑ. half of the original value
ⓒ. the same as the original value
ⓓ. four times the original value
Correct Answer: four times the original value
Explanation: Kinetic energy is \(K=\frac{1}{2}mv^2\). If mass remains the same, kinetic energy is proportional to \(v^2\). When speed changes from \(v\) to \(2v\), the square changes from \(v^2\) to \((2v)^2=4v^2\). Therefore, the kinetic energy becomes four times its original value. The doubling of speed does not merely double kinetic energy because the speed term is squared.
107. A body moving east and the same body moving west with the same speed have
ⓐ. equal kinetic energies and opposite momenta
ⓑ. opposite kinetic energies and equal momenta
ⓒ. zero kinetic energy in both cases
ⓓ. kinetic energies that depend on the chosen east-west direction
Correct Answer: equal kinetic energies and opposite momenta
Explanation: Kinetic energy depends on speed, not on the direction of velocity. If the same body has the same speed in two opposite directions, \(K=\frac{1}{2}mv^2\) has the same value in both cases. Momentum depends on velocity, so reversing direction reverses momentum. This shows why kinetic energy is scalar while momentum is vector. Direction changes the vector quantity, but it does not make kinetic energy negative or opposite.
108. Study the table about kinetic energy.
| Row | Claim | Status |
|---|
| P | Kinetic energy depends on \(v^2\) | true |
| Q | Kinetic energy is a scalar | true |
| R | Kinetic energy can be negative if velocity is negative | true |
| S | Kinetic energy is measured in \(\text{J}\) | true |
The row that needs correction is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row R
Explanation: Kinetic energy depends on \(v^2\), so the sign of velocity does not make kinetic energy negative. Row P is true because \(K=\frac{1}{2}mv^2\). Row Q is true because kinetic energy is a scalar quantity. Row S is true because kinetic energy is a form of energy and is measured in joule, \( \text{J} \). Row R is false because a negative velocity only indicates direction along a chosen axis, while \(v^2\) is non-negative.
109. A \(2\,\text{kg}\) body has kinetic energy \(100\,\text{J}\). Its speed is
ⓐ. \(5\,\text{m s}^{-1}\)
ⓑ. \(20\,\text{m s}^{-1}\)
ⓒ. \(50\,\text{m s}^{-1}\)
ⓓ. \(10\,\text{m s}^{-1}\)
Correct Answer: \(10\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given:} \) \(m=2\,\text{kg}\) and \(K=100\,\text{J}\).
\( \textbf{Required:} \) Speed \(v\).
Start with
\[
K=\frac{1}{2}mv^2
\]
Rearrange for \(v^2\):
\[
v^2=\frac{2K}{m}
\]
Substitute the values:
\[
v^2=\frac{2(100)}{2}
\]
\[
v^2=100
\]
Take the positive square root because speed is non-negative:
\[
v=10\,\text{m s}^{-1}
\]
The square-root step is necessary because kinetic energy contains \(v^2\).
\( \textbf{Final answer:} \) The speed is \(10\,\text{m s}^{-1}\).
110. Two bodies have the same speed. Body P has twice the mass of body Q. The kinetic energy of P is
ⓐ. half that of Q
ⓑ. equal to that of Q
ⓒ. twice that of Q
ⓓ. four times that of Q
Correct Answer: twice that of Q
Explanation: Kinetic energy is \(K=\frac{1}{2}mv^2\). If two bodies have the same speed, then \(v^2\) is the same for both. Kinetic energy is then directly proportional to mass. If body P has twice the mass of body Q, P has twice the kinetic energy of Q. This comparison uses mass proportionality only because the speed is kept unchanged.
111. A graph is plotted with kinetic energy \(K\) on the vertical axis and speed \(v\) on the horizontal axis for a fixed mass. The graph is
ⓐ. a straight line through the origin with constant slope
ⓑ. a horizontal straight line
ⓒ. a straight line with negative slope
ⓓ. a parabola opening upward
Correct Answer: a parabola opening upward
Explanation: For a fixed mass, kinetic energy is \(K=\frac{1}{2}mv^2\). This means \(K\) is proportional to \(v^2\), not directly proportional to \(v\). A relation of the form \(K\propto v^2\) gives a parabola opening upward when \(K\) is plotted against \(v\). The graph passes through the origin because zero speed gives zero kinetic energy. A straight-line graph would be expected for \(K\) versus \(v^2\), not for \(K\) versus \(v\).
112. Assertion: Kinetic energy of a moving body can never be negative.
Reason: Kinetic energy is proportional to the square of speed.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Kinetic energy is given by \(K=\frac{1}{2}mv^2\). Mass \(m\) is positive for an ordinary body, and speed squared \(v^2\) is never negative. Therefore, kinetic energy is zero when the body is at rest and positive when the body is moving. The Reason explains why a negative velocity along an axis does not produce negative kinetic energy. The sign of velocity belongs to direction, while kinetic energy depends on speed squared.
113. The work-energy theorem states that the net work done on a body is equal to
ⓐ. its final kinetic energy only
ⓑ. its change in potential energy only
ⓒ. the product of force and time
ⓓ. its change in kinetic energy
Correct Answer: its change in kinetic energy
Explanation: The work-energy theorem connects the net work done by all forces with the change in kinetic energy of a body. It is written as \(W_{\text{net}}=\Delta K=K_f-K_i\). The word net is important because the theorem uses the total work done by all forces, not just one selected force unless that is the only force doing work. If net work is positive, kinetic energy increases. If net work is negative, kinetic energy decreases. Force multiplied by time is related to impulse, not directly to change in kinetic energy.
114. A body has initial kinetic energy \(25\,\text{J}\). The net work done on it is \(40\,\text{J}\). Its final kinetic energy is
ⓐ. \(15\,\text{J}\)
ⓑ. \(25\,\text{J}\)
ⓒ. \(40\,\text{J}\)
ⓓ. \(65\,\text{J}\)
Correct Answer: \(65\,\text{J}\)
Explanation: \( \textbf{Given:} \) Initial kinetic energy \(K_i=25\,\text{J}\).
\( \textbf{Given:} \) Net work \(W_{\text{net}}=40\,\text{J}\).
\( \textbf{Work-energy theorem:} \)
\[
W_{\text{net}}=\Delta K=K_f-K_i
\]
\( \textbf{Rearrange for final kinetic energy:} \)
\[
K_f=K_i+W_{\text{net}}
\]
\( \textbf{Substitution:} \)
\[
K_f=25+40
\]
\[
K_f=65\,\text{J}
\]
The work is positive, so the kinetic energy must increase rather than decrease.
\( \textbf{Final answer:} \) The final kinetic energy is \(65\,\text{J}\).
115. A constant net force acts on a body of mass \(m\), changing its speed from \(u\) to \(v\) over displacement \(s\). The derivation of \(W_{\text{net}}=\Delta K\) uses \(F_{\text{net}}=ma\) along with
ⓐ. \(v=u+at\)
ⓑ. \(s=ut+\frac{1}{2}at^2\)
ⓒ. \(v^2-u^2=2as\)
ⓓ. \(p=mv\)
Correct Answer: \(v^2-u^2=2as\)
Explanation: For constant acceleration along the displacement, the kinematic relation \(v^2-u^2=2as\) directly connects speeds, acceleration, and displacement. Multiplying both sides suitably by \(m\) leads to \(\frac{1}{2}mv^2-\frac{1}{2}mu^2=mas\). Since \(F_{\text{net}}=ma\), the term \(mas\) becomes \(F_{\text{net}}s\), which is net work for force along displacement. This gives \(W_{\text{net}}=K_f-K_i\). The relation \(v=u+at\) contains time, but the work-energy theorem is naturally displacement and speed based in this derivation.
116. A \(2\,\text{kg}\) block initially at rest is acted on by a constant net force of \(10\,\text{N}\) over a displacement of \(5\,\text{m}\) on a horizontal line. Its final speed is
ⓐ. \(5\,\text{m s}^{-1}\)
ⓑ. \(10\,\text{m s}^{-1}\)
ⓒ. \(25\,\text{m s}^{-1}\)
ⓓ. \(7.1\,\text{m s}^{-1}\)
Correct Answer: \(7.1\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given:} \) \(m=2\,\text{kg}\), \(F_{\text{net}}=10\,\text{N}\), and \(s=5\,\text{m}\).
The block starts from rest, so \(K_i=0\).
\( \textbf{Net work:} \)
\[
W_{\text{net}}=F_{\text{net}}s
\]
\[
W_{\text{net}}=(10)(5)=50\,\text{J}
\]
\( \textbf{Work-energy theorem:} \)
\[
W_{\text{net}}=\Delta K=K_f-K_i
\]
Since \(K_i=0\),
\[
K_f=50\,\text{J}
\]
Use kinetic energy:
\[
K_f=\frac{1}{2}mv^2
\]
\[
50=\frac{1}{2}(2)v^2
\]
\[
50=v^2
\]
\[
v=\sqrt{50}\approx7.1\,\text{m s}^{-1}
\]
\( \textbf{Final answer:} \) The final speed is approximately \(7.1\,\text{m s}^{-1}\).
117. A car moving along a straight road has its kinetic energy reduced from \(5000\,\text{J}\) to \(1200\,\text{J}\) by braking forces. The net work done on the car is
ⓐ. \(+6200\,\text{J}\)
ⓑ. \(-6200\,\text{J}\)
ⓒ. \(+3800\,\text{J}\)
ⓓ. \(-3800\,\text{J}\)
Correct Answer: \(-3800\,\text{J}\)
Explanation: \( \textbf{Initial kinetic energy:} \) \(K_i=5000\,\text{J}\).
\( \textbf{Final kinetic energy:} \) \(K_f=1200\,\text{J}\).
By the work-energy theorem,
\[
W_{\text{net}}=\Delta K=K_f-K_i
\]
Substitute:
\[
W_{\text{net}}=1200-5000
\]
\[
W_{\text{net}}=-3800\,\text{J}
\]
The negative sign shows that the net force has reduced the car's kinetic energy.
Adding the two kinetic energies would ignore the meaning of change.
\( \textbf{Final answer:} \) The net work done on the car is \(-3800\,\text{J}\).
118. A body is moving to the right. A net force to the left acts on it for some displacement to the right. According to the work-energy theorem, the body's kinetic energy
ⓐ. increases because a force is present
ⓑ. decreases because the net work is negative
ⓒ. remains unchanged because force and displacement are both non-zero
ⓓ. becomes negative
Correct Answer: decreases because the net work is negative
Explanation: If the body is displaced to the right while the net force acts to the left, the angle between net force and displacement is \(180^\circ\). Therefore, the net work is negative. From \(W_{\text{net}}=\Delta K\), negative net work means negative change in kinetic energy. The body's kinetic energy decreases, usually appearing as a reduction in speed. Kinetic energy itself does not become negative; it can decrease down to zero but not below zero for ordinary motion.
119. The following statements are about the work-energy theorem.
I. It uses net work, not necessarily the work of a single chosen force.
II. Positive net work increases kinetic energy.
III. Zero net work always means the body is at rest.
IV. Negative net work decreases kinetic energy.
The true statements are
ⓐ. I, II, and IV only
ⓑ. I and III only
ⓒ. II, III, and IV only
ⓓ. I, II, III, and IV
Correct Answer: I, II, and IV only
Explanation: Statement I is true because the theorem is \(W_{\text{net}}=\Delta K\). Statement II is true because positive net work makes \(K_f-K_i\) positive. Statement IV is true because negative net work makes \(K_f-K_i\) negative. Statement III is false because zero net work means kinetic energy does not change, not necessarily that the body is at rest. A body moving at constant speed can have zero net work while still having non-zero kinetic energy.
120. A block of mass \(4\,\text{kg}\) moving at \(3\,\text{m s}^{-1}\) is acted on by a constant net force along its motion through \(2\,\text{m}\). If the force is \(16\,\text{N}\), its final speed is
ⓐ. \(4\,\text{m s}^{-1}\)
ⓑ. \(6\,\text{m s}^{-1}\)
ⓒ. \(7\,\text{m s}^{-1}\)
ⓓ. \(5\,\text{m s}^{-1}\)
Correct Answer: \(5\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given:} \) \(m=4\,\text{kg}\), \(u=3\,\text{m s}^{-1}\), \(F_{\text{net}}=16\,\text{N}\), and \(s=2\,\text{m}\).
\( \textbf{Initial kinetic energy:} \)
\[
K_i=\frac{1}{2}mu^2
\]
\[
K_i=\frac{1}{2}(4)(3)^2=2\times9=18\,\text{J}
\]
\( \textbf{Net work:} \)
\[
W_{\text{net}}=F_{\text{net}}s
\]
\[
W_{\text{net}}=(16)(2)=32\,\text{J}
\]
\( \textbf{Work-energy theorem:} \)
\[
K_f=K_i+W_{\text{net}}
\]
\[
K_f=18+32=50\,\text{J}
\]
Now use
\[
K_f=\frac{1}{2}mv^2
\]
\[
50=\frac{1}{2}(4)v^2=2v^2
\]
\[
v^2=25
\]
\[
v=5\,\text{m s}^{-1}
\]
Positive net work increases the kinetic energy from \(18\,\text{J}\) to \(50\,\text{J}\).
\( \textbf{Final answer:} \) The final speed is \(5\,\text{m s}^{-1}\).
Discussing about Que no:
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∘ with the direction of motion, what is the work done?
GM Sir/Madam,
I think, the options and the key given for the question no. 103 is wrong. The correct answer is 69.2. Please check it once. Thank you.
Hi Gutti.Akshitha!
Big thanks for spotting that mistake in Question 103. You were right on the mark—the correct answer is 69.2 J.
We’ve just updated the question and the answer key on the site. We really appreciate you taking the time to report this and help us improve. Keep up the great study work!
Cheers, Team gkaim