401. A spring is stretched slightly and then released. The spring force on the attached body is called restoring because it acts
ⓐ. opposite to the displacement
ⓑ. in the direction of displacement from natural length
ⓒ. vertically downward in every case
ⓓ. only when the body has zero mass
Correct Answer: opposite to the displacement
Explanation: A spring force tries to bring the spring back toward its natural length. If the spring is stretched, it pulls the attached body back toward the unstretched position. If the spring is compressed, it pushes the attached body away from the compressed side toward the natural length. This is why the spring force is called a restoring force. In one-dimensional form, Hooke’s law is often written as \(F=-kx\), where the negative sign shows that force and displacement from equilibrium are opposite in direction. The force is not always downward; its direction depends on how the spring is placed and deformed.
402. In Hooke’s law \(F=-kx\), the SI unit of spring constant \(k\) is
ⓐ. \( \text{N m} \)
ⓑ. \( \text{kg m s}^{-1} \)
ⓒ. \( \text{N m}^{-1} \)
ⓓ. \( \text{m N}^{-1} \)
Correct Answer: \( \text{N m}^{-1} \)
Explanation: Hooke’s law relates spring force to displacement by \(F=-kx\). The magnitude form is \(F=kx\). Therefore \(k=\frac{F}{x}\). The SI unit of force is \( \text{N} \), and the SI unit of displacement is \( \text{m} \). Hence the unit of \(k\) is \( \text{N m}^{-1} \). The unit \( \text{N m} \) is associated with work or torque, not spring constant. A larger \(k\) means a larger force is needed for the same extension or compression.
403. A spring of constant \(200\,\text{N m}^{-1}\) is stretched by \(5.0\,\text{cm}\). The magnitude of the restoring force is
ⓐ. \(40\,\text{N}\)
ⓑ. \(1000\,\text{N}\)
ⓒ. \(4\,\text{N}\)
ⓓ. \(10\,\text{N}\)
Correct Answer: \(10\,\text{N}\)
Explanation: \( \textbf{Given:} \) Spring constant \(k=200\,\text{N m}^{-1}\).
\( \textbf{Extension:} \)
\[x=5.0\,\text{cm}=0.050\,\text{m}\]
For magnitude of spring force:
\[F=kx\]
\( \textbf{Substitution:} \)
\[F=(200\,\text{N m}^{-1})(0.050\,\text{m})\]
\( \textbf{Calculation:} \)
\[F=10\,\text{N}\]
The restoring force acts opposite to the extension direction, but the question asks for magnitude.
The centimetre-to-metre conversion is necessary because \(k\) is given in \( \text{N m}^{-1} \).
\( \textbf{Final answer:} \) The restoring force has magnitude \(10\,\text{N}\).
404. Hooke’s law is most safely used when the spring
ⓐ. within its elastic limit
ⓑ. is stretched to any length without limit
ⓒ. has no restoring tendency
ⓓ. is disconnected from all forces
Correct Answer: within its elastic limit
Explanation: Hooke’s law states that the restoring force of a spring is proportional to displacement from natural length. This simple proportional relation is valid only within the elastic limit of the spring. If the spring is stretched or compressed too much, the relation may no longer remain linear. The spring may also suffer permanent deformation beyond the elastic limit. The law does not say that any amount of extension is allowed. The condition of small elastic deformation is part of using \(F=-kx\) correctly.
405. A \(3\,\text{kg}\) block is pulled on a smooth horizontal surface by a horizontal force of \(12\,\text{N}\). Its acceleration is
ⓐ. \(3\,\text{m s}^{-2}\)
ⓑ. \(12\,\text{m s}^{-2}\)
ⓒ. \(4\,\text{m s}^{-2}\)
ⓓ. \(36\,\text{m s}^{-2}\)
Correct Answer: \(4\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given:} \) Mass \(m=3\,\text{kg}\) and horizontal force \(F=12\,\text{N}\).
The surface is smooth, so friction is absent.
The vertical forces \(N\) and \(mg\) balance because there is no vertical acceleration.
\( \textbf{Horizontal equation:} \)
\[\sum F_x=ma_x\]
\( \textbf{Substitution:} \)
\[12=(3)a_x\]
\( \textbf{Solving:} \)
\[a_x=\frac{12}{3}=4\,\text{m s}^{-2}\]
The acceleration is in the direction of the applied horizontal force.
\( \textbf{Final answer:} \) The acceleration is \(4\,\text{m s}^{-2}\).
406. Two blocks of masses \(2\,\text{kg}\) and \(3\,\text{kg}\) are connected by a light string on a smooth horizontal surface. A horizontal force of \(10\,\text{N}\) pulls the \(3\,\text{kg}\) block. The common acceleration of the system is
ⓐ. \(5\,\text{m s}^{-2}\)
ⓑ. \(2\,\text{m s}^{-2}\)
ⓒ. \(3.3\,\text{m s}^{-2}\)
ⓓ. \(10\,\text{m s}^{-2}\)
Correct Answer: \(2\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given:} \) \(m_1=2\,\text{kg}\), \(m_2=3\,\text{kg}\), and external pull \(F=10\,\text{N}\).
The surface is smooth, so there is no friction.
Treat the two blocks together as one system.
The string tension is internal to the two-block system.
\( \textbf{Total mass:} \)
\[m_{\text{total}}=2+3=5\,\text{kg}\]
\( \textbf{System equation:} \)
\[F=m_{\text{total}}a\]
\( \textbf{Substitution:} \)
\[10=(5)a\]
\[a=2\,\text{m s}^{-2}\]
Using the total mass is valid because both blocks have the same acceleration.
\( \textbf{Final answer:} \) The common acceleration is \(2\,\text{m s}^{-2}\).
407. In the connected-block system on a smooth horizontal surface, the tension in the string is best found by
ⓐ. treating tension as the only external force on the two-block system
ⓑ. applying \(F=ma\) to one block after finding common acceleration
ⓒ. setting tension equal to the total applied force always
ⓓ. ignoring the string because it is light
Correct Answer: applying \(F=ma\) to one block after finding common acceleration
Explanation: When both connected blocks are treated as one system, the string tension is internal and cancels from the system equation. This is useful for finding the common acceleration. To find tension, one block must then be isolated and Newton’s second law applied to that block. For the block not directly pulled by the external force, tension may be the horizontal force causing its acceleration. A light string can still exert tension; “light” means its mass is neglected. Tension is not automatically equal to the applied force because part of the applied force accelerates the whole system.
408. Two blocks of masses \(2\,\text{kg}\) and \(3\,\text{kg}\) are connected by a light string on a smooth horizontal surface. A \(10\,\text{N}\) force pulls the \(3\,\text{kg}\) block, and the common acceleration is \(2\,\text{m s}^{-2}\). The tension in the string is
ⓐ. \(6\,\text{N}\)
ⓑ. \(4\,\text{N}\)
ⓒ. \(10\,\text{N}\)
ⓓ. \(2\,\text{N}\)
Correct Answer: \(4\,\text{N}\)
Explanation: \( \textbf{Given:} \) \(m_1=2\,\text{kg}\), \(m_2=3\,\text{kg}\), and common acceleration \(a=2\,\text{m s}^{-2}\).
The \(3\,\text{kg}\) block is directly pulled, while the \(2\,\text{kg}\) block is pulled by the string.
To find tension, isolate the \(2\,\text{kg}\) block.
\( \textbf{Horizontal force on the \(2\,\text{kg}\) block:} \) tension \(T\).
\( \textbf{Newton’s second law for that block:} \)
\[T=m_1a\]
\( \textbf{Substitution:} \)
\[T=(2)(2)\]
\[T=4\,\text{N}\]
The applied force is \(10\,\text{N}\), but only \(4\,\text{N}\) is needed to accelerate the \(2\,\text{kg}\) block.
\( \textbf{Final answer:} \) The tension is \(4\,\text{N}\).
409. The table describes a two-block horizontal system on a smooth surface.
| Row | Statement |
|---|
| P | Both connected blocks have the same acceleration. |
| Q | The string tension is internal if both blocks are chosen as one system. |
| R | The string tension is external if only one block is chosen as the system. |
| S | The tension must always equal the applied pulling force. |
The unsuitable row is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Row P is suitable because an ideal string constrains the connected blocks to have a common acceleration. Row Q is suitable because tension is an interaction between the two blocks through the string when both blocks are taken as one system. Row R is also suitable because, for one selected block, the string force comes from outside that one-block system. Row S is unsuitable because the applied pulling force is usually shared between accelerating the system and producing tension. The value of tension depends on which block is examined and what mass it must accelerate.
410. A force \(F\) pulls two connected blocks of masses \(m_1\) and \(m_2\) on a smooth horizontal surface. If \(F\) is applied to \(m_2\), the tension in the string is
ⓐ. \(\frac{m_2F}{m_1+m_2}\)
ⓑ. \(F\)
ⓒ. \(\frac{m_1F}{m_1+m_2}\)
ⓓ. \((m_1+m_2)F\)
Correct Answer: \(\frac{m_1F}{m_1+m_2}\)
Explanation: Treat both blocks as one system first.
The total mass is \(m_1+m_2\).
The common acceleration is:
\[a=\frac{F}{m_1+m_2}\]
The block \(m_1\) is pulled only by the string tension \(T\).
For block \(m_1\):
\[T=m_1a\]
Substitute the acceleration:
\[T=m_1\left(\frac{F}{m_1+m_2}\right)\]
So:
\[T=\frac{m_1F}{m_1+m_2}\]
The tension depends on the mass that the string has to pull, not just on the total applied force.
411. In an ideal Atwood machine, two masses are connected by a light string passing over a smooth pulley. The heavier mass moves downward because
ⓐ. the tension is zero in the string
ⓑ. both masses must move downward together
ⓒ. the lighter mass has no weight
ⓓ. heavier weight is larger
Correct Answer: heavier weight is larger
Explanation: In an Atwood machine, the two masses are connected, so they have accelerations of equal magnitude in opposite directions. If one mass is heavier, its weight is larger. The difference in the two weights provides the net driving force for the two-mass system. Tension is not zero; it transmits the interaction through the string. The lighter mass moves upward while the heavier mass moves downward. The motion is determined by the imbalance of weights, not by the disappearance of force on either mass.
412. Two masses \(m_1\) and \(m_2\) are connected over an ideal pulley, with \(m_2\gt m_1\). The acceleration magnitude of the system is
ⓐ. \(\frac{(m_2-m_1)g}{m_1+m_2}\)
ⓑ. \(\frac{(m_1+m_2)g}{m_2-m_1}\)
ⓒ. \((m_2-m_1)g\)
ⓓ. \(\frac{m_1g}{m_2}\)
Correct Answer: \(\frac{(m_2-m_1)g}{m_1+m_2}\)
Explanation: Take \(m_2\) moving downward and \(m_1\) moving upward.
For \(m_2\):
\[m_2g-T=m_2a\]
For \(m_1\):
\[T-m_1g=m_1a\]
Add the two equations:
\[m_2g-m_1g=(m_1+m_2)a\]
Factor \(g\):
\[(m_2-m_1)g=(m_1+m_2)a\]
Solving for \(a\):
\[a=\frac{(m_2-m_1)g}{m_1+m_2}\]
The acceleration becomes zero when \(m_1=m_2\), as expected for balanced weights.
\( \textbf{Final answer:} \) \(a=\frac{(m_2-m_1)g}{m_1+m_2}\).
413. In an Atwood machine, \(m_1=2\,\text{kg}\), \(m_2=3\,\text{kg}\), and \(g=10\,\text{m s}^{-2}\). The magnitude of acceleration is
ⓐ. \(2\,\text{m s}^{-2}\)
ⓑ. \(5\,\text{m s}^{-2}\)
ⓒ. \(1\,\text{m s}^{-2}\)
ⓓ. \(10\,\text{m s}^{-2}\)
Correct Answer: \(2\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given:} \) \(m_1=2\,\text{kg}\), \(m_2=3\,\text{kg}\), and \(g=10\,\text{m s}^{-2}\).
Since \(m_2\gt m_1\), \(m_2\) moves downward.
\( \textbf{Atwood acceleration:} \)
\[a=\frac{(m_2-m_1)g}{m_1+m_2}\]
\( \textbf{Substitution:} \)
\[a=\frac{(3-2)(10)}{2+3}\]
\( \textbf{Calculation:} \)
\[a=\frac{10}{5}=2\,\text{m s}^{-2}\]
The acceleration is less than \(g\) because the lighter mass and string tension oppose the motion of the heavier mass.
\( \textbf{Final answer:} \) The acceleration magnitude is \(2\,\text{m s}^{-2}\).
414. For the Atwood machine with \(m_1=2\,\text{kg}\), \(m_2=3\,\text{kg}\), \(g=10\,\text{m s}^{-2}\), and acceleration \(2\,\text{m s}^{-2}\), the tension in the string is
ⓐ. \(20\,\text{N}\)
ⓑ. \(30\,\text{N}\)
ⓒ. \(24\,\text{N}\)
ⓓ. \(12\,\text{N}\)
Correct Answer: \(24\,\text{N}\)
Explanation: \( \textbf{Given:} \) \(m_1=2\,\text{kg}\), \(m_2=3\,\text{kg}\), \(a=2\,\text{m s}^{-2}\), and \(g=10\,\text{m s}^{-2}\).
The lighter mass \(m_1\) moves upward.
For \(m_1\), tension acts upward and weight acts downward.
\( \textbf{Equation for \(m_1\):} \)
\[T-m_1g=m_1a\]
\( \textbf{Solving for tension:} \)
\[T=m_1(g+a)\]
\( \textbf{Substitution:} \)
\[T=2(10+2)\]
\[T=24\,\text{N}\]
The same value can be obtained from \(m_2g-T=m_2a\).
The tension lies between the two weights \(20\,\text{N}\) and \(30\,\text{N}\).
\( \textbf{Final answer:} \) The tension is \(24\,\text{N}\).
415. A passage about a pulley system is given below.
Two unequal masses are connected by a light string over a smooth pulley. A student writes one equation for each mass but chooses downward as positive for both masses.
What is the main issue with this setup?
ⓐ. A smooth pulley makes the acceleration zero
ⓑ. Tension cannot act in a string over a pulley
ⓒ. Gravity acts upward on the lighter mass
ⓓ. the acceleration signs must be opposite
Correct Answer: the acceleration signs must be opposite
Explanation: In an ideal Atwood machine, one mass moves downward while the other moves upward. Their acceleration magnitudes are equal, but their directions are opposite. If the same positive direction is chosen for both sides without care, one acceleration should carry a negative sign. A simpler method is to choose positive along the expected motion for each mass separately and write both equations with the same acceleration magnitude. The smooth pulley does not make acceleration zero; it only allows the same tension in the ideal string. Correct signs are essential because the string constraint connects the two motions.
416. The table lists proposed equations for an Atwood machine with \(m_2\gt m_1\), where \(m_2\) moves down and \(m_1\) moves up.
| Row | Equation for \(m_2\) | Equation for \(m_1\) |
|---|
| P | \(m_2g-T=m_2a\) | \(T-m_1g=m_1a\) |
| Q | \(T-m_2g=m_2a\) | \(T-m_1g=m_1a\) |
| R | \(m_2g+T=m_2a\) | \(m_1g+T=m_1a\) |
| S | \(m_2g-T=0\) | \(T-m_1g=0\) |
The suitable row is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: Since \(m_2\) moves downward, choose downward as positive for \(m_2\). Its weight \(m_2g\) is positive and tension \(T\) is upward, so \(m_2g-T=m_2a\). Since \(m_1\) moves upward, choose upward as positive for \(m_1\). Its tension \(T\) is positive and weight \(m_1g\) is downward, so \(T-m_1g=m_1a\). Row P uses these directions consistently. Rows Q and R give incorrect force directions, while row S incorrectly assumes zero acceleration. The sign convention should match the actual or assumed direction of motion on each side.
417. On a smooth inclined plane of angle \(\theta\), a block released from rest accelerates down the plane because the unbalanced component of weight along the plane is
ⓐ. \(mg\cos\theta\)
ⓑ. \(mg\tan\theta\)
ⓒ. \(mg\sin\theta\)
ⓓ. \(mg\)
Correct Answer: \(mg\sin\theta\)
Explanation: The weight \(mg\) acts vertically downward. On an inclined plane, it is resolved into a component perpendicular to the plane and a component parallel to the plane. The perpendicular component is \(mg\cos\theta\), and it is balanced by the normal reaction on a smooth plane. The component along the plane is \(mg\sin\theta\), directed down the incline. Since there is no friction, this down-plane component is the net force along the plane. This is the force responsible for the block’s acceleration.
418. A block slides down a smooth inclined plane of angle \(30^\circ\). Taking \(g=10\,\text{m s}^{-2}\), its acceleration down the plane is
ⓐ. \(10\,\text{m s}^{-2}\)
ⓑ. \(8.66\,\text{m s}^{-2}\)
ⓒ. \(2.5\,\text{m s}^{-2}\)
ⓓ. \(5\,\text{m s}^{-2}\)
Correct Answer: \(5\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given:} \) Incline angle \(\theta=30^\circ\) and \(g=10\,\text{m s}^{-2}\).
The plane is smooth, so friction is absent.
\( \textbf{Down-plane force:} \)
\[\sum F=mg\sin\theta\]
\( \textbf{Newton’s second law along the plane:} \)
\[ma=mg\sin\theta\]
Cancel \(m\):
\[a=g\sin\theta\]
\( \textbf{Substitution:} \)
\[a=10\sin30^\circ\]
\[\sin30^\circ=\frac{1}{2}\]
\[a=10\left(\frac{1}{2}\right)=5\,\text{m s}^{-2}\]
The mass cancels, so all blocks slide with the same acceleration on the same smooth incline.
\( \textbf{Final answer:} \) The acceleration is \(5\,\text{m s}^{-2}\).
419. A graph of acceleration \(a\) down a smooth inclined plane is plotted against \(\sin\theta\). The graph is a straight line through the origin. Its slope is
ⓐ. \(mg\)
ⓑ. \(\frac{1}{g}\)
ⓒ. \(m\)
ⓓ. \(g\)
Correct Answer: \(g\)
Explanation: For a smooth inclined plane:
\[a=g\sin\theta\]
If \(a\) is plotted on the vertical axis and \(\sin\theta\) on the horizontal axis, the equation has the form \(y=(\text{slope})x\). The slope is therefore \(g\). The graph passes through the origin because \(a=0\) when \(\theta=0^\circ\). The mass does not appear because it cancels from \(ma=mg\sin\theta\). This graph is a clean way to see the proportionality between down-plane acceleration and \(\sin\theta\).
420. A block of mass \(m\) is on a smooth inclined plane of angle \(\theta\) and is connected by a light string over a smooth pulley to a hanging mass \(M\). If the hanging mass moves downward, the correct system acceleration is
ⓐ. \(\frac{Mg+mg\sin\theta}{M+m}\)
ⓑ. \(\frac{mg\cos\theta-Mg}{M+m}\)
ⓒ. \(\frac{Mg-mg\sin\theta}{M+m}\)
ⓓ. \(\frac{Mg}{M}\)
Correct Answer: \(\frac{Mg-mg\sin\theta}{M+m}\)
Explanation: The hanging mass \(M\) moves downward, so its weight \(Mg\) drives the motion.
The block on the incline moves up the plane.
For the hanging mass:
\[Mg-T=Ma\]
For the block on the incline:
\[T-mg\sin\theta=ma\]
Add the two equations:
\[Mg-mg\sin\theta=(M+m)a\]
Solve for \(a\):
\[a=\frac{Mg-mg\sin\theta}{M+m}\]
This expression is positive only if the assumed direction is correct.
If the numerator is negative, the actual motion is opposite to the assumed direction.
\( \textbf{Final answer:} \) \(a=\frac{Mg-mg\sin\theta}{M+m}\).
