201. Assertion: A particle placed exactly at a local maximum of \(U(x)\) can be in equilibrium.
Reason: At a local maximum, the slope of \(U-x\) graph is zero.
ⓐ. 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: Equilibrium requires the force to be zero. For a conservative force in one dimension, \(F=-\frac{dU}{dx}\). At a local maximum of \(U(x)\), the tangent to the curve is horizontal, so \(\frac{dU}{dx}=0\). Hence, \(F=0\) at that exact point, and the particle can be in equilibrium. However, this equilibrium is unstable because a small displacement on either side makes the particle move away toward lower potential energy.
202. In a region where the \(U-x\) graph is perfectly horizontal, the force on the particle is
ⓐ. positive and constant
ⓑ. negative and constant
ⓒ. zero
ⓓ. infinite
Correct Answer: zero
Explanation: A horizontal \(U-x\) graph has zero slope. Since the force is \(F(x)=-\frac{dU}{dx}\), zero slope gives zero force. This does not require the potential energy itself to be zero; the potential energy may have any constant value in that region. Force is connected to change of potential energy with position, not to the absolute value of potential energy. A flat potential-energy region means there is no tendency toward increasing or decreasing \(x\) due to that conservative force.
203. For a particle moving in one dimension with total mechanical energy \(E\), the kinetic energy at position \(x\) is
ⓐ. \(K=U-E\)
ⓑ. \(K=EU\)
ⓒ. \(K=\frac{U}{E}\)
ⓓ. \(K=E-U\)
Correct Answer: \(K=E-U\)
Explanation: Mechanical energy is the sum of kinetic and potential energies, \(E=K+U\). Rearranging this relation gives \(K=E-U\). Since kinetic energy cannot be negative, motion is allowed only where \(E-U\geq0\). This means the total energy line must lie at or above the potential energy curve. Turning points occur where \(E=U\), because \(K=0\) there.
204. A particle has total mechanical energy \(50\,\text{J}\). At a certain position, its potential energy is \(18\,\text{J}\). Its kinetic energy at that position is
ⓐ. \(18\,\text{J}\)
ⓑ. \(32\,\text{J}\)
ⓒ. \(50\,\text{J}\)
ⓓ. \(68\,\text{J}\)
Correct Answer: \(32\,\text{J}\)
Explanation: \( \textbf{Given:} \) Total mechanical energy \(E=50\,\text{J}\).
\( \textbf{Given:} \) Potential energy at the position \(U=18\,\text{J}\).
Mechanical energy is
\[
E=K+U
\]
Rearrange:
\[
K=E-U
\]
Substitute:
\[
K=50-18
\]
\[
K=32\,\text{J}
\]
The kinetic energy is the part of total mechanical energy left after accounting for potential energy.
\( \textbf{Final answer:} \) The kinetic energy is \(32\,\text{J}\).
205. A particle of mass \(2\,\text{kg}\) has total mechanical energy \(40\,\text{J}\). At a point where \(U=24\,\text{J}\), its speed is
ⓐ. \(2\,\text{m s}^{-1}\)
ⓑ. \(6\,\text{m s}^{-1}\)
ⓒ. \(8\,\text{m s}^{-1}\)
ⓓ. \(4\,\text{m s}^{-1}\)
Correct Answer: \(4\,\text{m s}^{-1}\)
Explanation: \( \textbf{Total energy:} \) \(E=40\,\text{J}\).
\( \textbf{Potential energy:} \) \(U=24\,\text{J}\).
\( \textbf{Mass:} \) \(m=2\,\text{kg}\).
First find kinetic energy:
\[
K=E-U
\]
\[
K=40-24=16\,\text{J}
\]
Use
\[
K=\frac{1}{2}mv^2
\]
Substitute:
\[
16=\frac{1}{2}(2)v^2
\]
\[
16=v^2
\]
\[
v=4\,\text{m s}^{-1}
\]
The speed is found from the energy difference, not from the value of \(U\) alone.
\( \textbf{Final answer:} \) The speed is \(4\,\text{m s}^{-1}\).
206. A point on a potential energy curve is called a turning point of the motion when
ⓐ. \(K\) is maximum
ⓑ. \(U=0\)
ⓒ. \(K=0\) and \(E=U\)
ⓓ. \(F=0\) always
Correct Answer: \(K=0\) and \(E=U\)
Explanation: At a turning point, the particle momentarily stops before reversing direction. Therefore, its kinetic energy is zero at that point. Since total mechanical energy is \(E=K+U\), setting \(K=0\) gives \(E=U\). This is why turning points are found where the total energy line meets the potential energy curve. A turning point is not necessarily an equilibrium point, because the force there need not be zero.
207. Use the graph description below.
A \(U-x\) graph has a valley shape. A horizontal total-energy line \(E\) cuts the curve at two points, labelled P and Q. Between P and Q, the curve lies below the energy line. Outside P and Q, the curve lies above the energy line.
The allowed region of motion is
ⓐ. only outside P and Q
ⓑ. only at P and Q
ⓒ. nowhere, because the curve is not horizontal
ⓓ. only between P and Q
Correct Answer: only between P and Q
Explanation: The kinetic energy is \(K=E-U\). Motion is possible only where \(K\geq0\), so the total energy \(E\) must be greater than or equal to \(U\). Between P and Q, the potential energy curve lies below the total-energy line, so \(E\geq U\). At P and Q, \(E=U\), so the kinetic energy is zero and these points are turning points. Outside P and Q, \(U\) is greater than \(E\), which would require negative kinetic energy and is not physically allowed.
208. Study the table for a particle with fixed total energy \(E=30\,\text{J}\).
| Position | Potential energy \(U\) | Motion status |
|---|
| P | \(10\,\text{J}\) | allowed |
| Q | \(30\,\text{J}\) | turning point possible |
| R | \(45\,\text{J}\) | allowed |
| S | \(25\,\text{J}\) | allowed |
The row that needs correction is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row R
Explanation: A position is allowed only if \(E\geq U\). At P, \(E=30\,\text{J}\) is greater than \(U=10\,\text{J}\), so motion is allowed. At Q, \(E=U=30\,\text{J}\), so \(K=0\), making it a possible turning point. At S, \(E=30\,\text{J}\) is greater than \(U=25\,\text{J}\), so motion is allowed. Row R is incorrect because \(U=45\,\text{J}\) is greater than \(E=30\,\text{J}\), which would require negative kinetic energy.
209. A particle of mass \(1\,\text{kg}\) moves in a potential-energy field with total energy \(20\,\text{J}\). At position P, \(U=12\,\text{J}\); at position Q, \(U=5\,\text{J}\). The speed is greater at
ⓐ. P, because \(U\) is larger there
ⓑ. both positions equally
ⓒ. neither position, because total energy is fixed
ⓓ. Q, because \(K=E-U\) is larger there
Correct Answer: Q, because \(K=E-U\) is larger there
Explanation: The total mechanical energy is fixed at \(E=20\,\text{J}\). At P, the kinetic energy is \(K_P=20-12=8\,\text{J}\). At Q, the kinetic energy is \(K_Q=20-5=15\,\text{J}\). Since the mass is the same, greater kinetic energy means greater speed. A lower potential energy point allows more of the fixed total energy to appear as kinetic energy.
210. A particle is moving in a potential-energy curve with fixed total energy. The statements below are made:
I. Motion is allowed where \(E\geq U\).
II. At a turning point, \(K=0\).
III. The particle is fastest where \(U\) is lowest within the allowed region.
IV. Motion is allowed where \(U>E\).
The true statements are
ⓐ. I, II, and III only
ⓑ. I and IV only
ⓒ. II, III, and IV only
ⓓ. I, II, III, and IV
Correct Answer: I, II, and III only
Explanation: Since \(E=K+U\), the kinetic energy is \(K=E-U\). Kinetic energy cannot be negative, so motion is allowed only where \(E\geq U\), making statement I true. At a turning point, \(E=U\), so \(K=0\), making statement II true. For fixed total energy and mass, the speed is greatest where \(K\) is greatest, which occurs where \(U\) is lowest in the allowed region, so statement III is true. Statement IV is false because \(U>E\) would make \(K\) negative, which is not possible for ordinary motion.
211. Mechanical energy of a system is conserved when
ⓐ. only conservative forces do work
ⓑ. friction does negative work throughout the motion
ⓒ. the body remains at rest throughout the motion
ⓓ. potential energy is zero at every point
Correct Answer: only conservative forces do work
Explanation: Mechanical energy is \(E=K+U\). It remains constant when the energy only changes between kinetic and potential forms through conservative forces. If friction or another non-conservative force does work, mechanical energy may decrease or increase depending on the situation. Conservation of mechanical energy does not require the body to be at rest; it often describes moving systems. The condition depends on the forces and the chosen system, not on the numerical value chosen for potential energy.
212. A block moves on a smooth track under gravity only. At one point, \(K=20\,\text{J}\) and \(U=50\,\text{J}\). At another point, \(U=35\,\text{J}\). Its kinetic energy there is
ⓐ. \(15\,\text{J}\)
ⓑ. \(55\,\text{J}\)
ⓒ. \(70\,\text{J}\)
ⓓ. \(35\,\text{J}\)
Correct Answer: \(35\,\text{J}\)
Explanation: \( \textbf{Initial energy data:} \) \(K_1=20\,\text{J}\) and \(U_1=50\,\text{J}\).
Mechanical energy is
\[
E=K+U
\]
So at the first point,
\[
E=20+50=70\,\text{J}
\]
The track is smooth and gravity is conservative, so mechanical energy remains constant.
At the second point,
\[
E=K_2+U_2
\]
\[
70=K_2+35
\]
\[
K_2=35\,\text{J}
\]
The decrease in potential energy becomes an equal increase in kinetic energy.
\( \textbf{Final answer:} \) The kinetic energy at the second point is \(35\,\text{J}\).
213. A stone is dropped from rest from height \(h\) above the ground, neglecting air resistance. Just before reaching the ground, its speed is
ⓐ. \(\sqrt{gh}\)
ⓑ. \(\sqrt{2gh}\)
ⓒ. \(2gh\)
ⓓ. \(\frac{h}{g}\)
Correct Answer: \(\sqrt{2gh}\)
Explanation: \( \textbf{Initial condition:} \) The stone is dropped from rest, so \(K_i=0\).
Choose the ground as the zero level of gravitational potential energy.
At the top,
\[
U_i=mgh
\]
Just before reaching the ground,
\[
U_f=0
\]
With no air resistance, mechanical energy is conserved:
\[
K_i+U_i=K_f+U_f
\]
\[
0+mgh=\frac{1}{2}mv^2+0
\]
Cancel \(m\):
\[
gh=\frac{1}{2}v^2
\]
\[
v^2=2gh
\]
\[
v=\sqrt{2gh}
\]
The result is independent of mass because both \(K\) and \(U\) contain the same factor \(m\).
\( \textbf{Final answer:} \) The speed is \(\sqrt{2gh}\).
214. A ball of mass \(0.5\,\text{kg}\) is dropped from rest from a height of \(20\,\text{m}\). Taking \(g=10\,\text{m s}^{-2}\) and neglecting air resistance, its speed just before hitting the ground is
ⓐ. \(10\,\text{m s}^{-1}\)
ⓑ. \(14\,\text{m s}^{-1}\)
ⓒ. \(20\,\text{m s}^{-1}\)
ⓓ. \(40\,\text{m s}^{-1}\)
Correct Answer: \(20\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given:} \) \(h=20\,\text{m}\) and \(g=10\,\text{m s}^{-2}\).
The ball is dropped from rest, so initial kinetic energy is zero.
With air resistance neglected, gravitational potential energy changes into kinetic energy.
\[
mgh=\frac{1}{2}mv^2
\]
Cancel the common factor \(m\):
\[
gh=\frac{1}{2}v^2
\]
\[
v^2=2gh
\]
Substitute values:
\[
v^2=2(10)(20)
\]
\[
v^2=400
\]
\[
v=20\,\text{m s}^{-1}
\]
The mass \(0.5\,\text{kg}\) cancels, so it does not affect the final speed in this ideal case.
\( \textbf{Final answer:} \) The speed just before hitting the ground is \(20\,\text{m s}^{-1}\).
215. A body is projected vertically upward with initial speed \(u\). Neglecting air resistance, the maximum height reached above the projection point is
ⓐ. \(\frac{u}{2g}\)
ⓑ. \(\frac{2g}{u^2}\)
ⓒ. \(2gu^2\)
ⓓ. \(\frac{u^2}{2g}\)
Correct Answer: \(\frac{u^2}{2g}\)
Explanation: At the projection point, take gravitational potential energy as zero.
The initial kinetic energy is
\[
K_i=\frac{1}{2}mu^2
\]
At maximum height, the speed is momentarily zero, so
\[
K_f=0
\]
If air resistance is neglected, mechanical energy is conserved:
\[
K_i+U_i=K_f+U_f
\]
\[
\frac{1}{2}mu^2+0=0+mgh
\]
Cancel \(m\):
\[
\frac{1}{2}u^2=gh
\]
Solve for height:
\[
h=\frac{u^2}{2g}
\]
The height depends on \(u^2\), so doubling the launch speed makes the maximum height four times.
\( \textbf{Final answer:} \) The maximum height is \(\frac{u^2}{2g}\).
216. A \(1\,\text{kg}\) object is thrown vertically upward with speed \(10\,\text{m s}^{-1}\). Taking \(g=10\,\text{m s}^{-2}\), its maximum height above the launch point is
ⓐ. \(2.5\,\text{m}\)
ⓑ. \(5.0\,\text{m}\)
ⓒ. \(10.0\,\text{m}\)
ⓓ. \(20.0\,\text{m}\)
Correct Answer: \(5.0\,\text{m}\)
Explanation: \( \textbf{Given:} \) \(u=10\,\text{m s}^{-1}\) and \(g=10\,\text{m s}^{-2}\).
At maximum height, the speed becomes zero.
Use energy conservation between launch point and highest point.
\[
\frac{1}{2}mu^2=mgh
\]
Cancel \(m\):
\[
\frac{1}{2}u^2=gh
\]
Solve for \(h\):
\[
h=\frac{u^2}{2g}
\]
Substitute:
\[
h=\frac{(10)^2}{2(10)}
\]
\[
h=\frac{100}{20}
\]
\[
h=5.0\,\text{m}
\]
The mass does not affect the height when air resistance is absent.
\( \textbf{Final answer:} \) The maximum height is \(5.0\,\text{m}\).
217. Two balls of different masses are dropped from the same height in vacuum. The reason they reach the ground with the same speed is that
ⓐ. heavier bodies have no gravitational potential energy
ⓑ. kinetic energy is independent of speed
ⓒ. gravitational acceleration becomes zero during falling
ⓓ. mass cancels in \(mgh=\frac{1}{2}mv^2\)
Correct Answer: mass cancels in \(mgh=\frac{1}{2}mv^2\)
Explanation: In vacuum, air resistance is absent and mechanical energy is conserved. The gravitational potential energy lost is \(mgh\). The kinetic energy gained just before impact is \(\frac{1}{2}mv^2\). Equating them gives \(mgh=\frac{1}{2}mv^2\), and the factor \(m\) cancels from both sides. This is why the final speed depends on height and \(g\), not on mass, in the ideal falling case.
218. Study the table about mechanical energy conservation.
| Row | Situation | Conclusion |
|---|
| P | Only gravity acts, air resistance neglected | \(K+U\) is constant |
| Q | Smooth spring-block system | \(K+U_s\) is constant |
| R | Rough horizontal surface with kinetic friction | \(K+U\) is necessarily constant |
| S | Smooth vertical motion under gravity | loss of \(U\) becomes gain of \(K\) |
The row that needs correction is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row R
Explanation: Row P is correct because gravity is conservative and air resistance is neglected. Row Q is correct because an ideal spring force is conservative and the surface is smooth. Row S is correct because in smooth vertical motion, gravitational potential energy and kinetic energy can transform into each other. Row R is incorrect because kinetic friction is non-conservative and usually reduces mechanical energy. On a rough surface, the energy accounting must include thermal/internal energy or non-conservative work.
219. A pendulum bob rises from the lowest point to a height \(0.20\,\text{m}\) above it. Neglecting air resistance, the speed it must have at the lowest point to just reach that height is
ⓐ. \(1.0\,\text{m s}^{-1}\)
ⓑ. \(4.0\,\text{m s}^{-1}\)
ⓒ. \(20\,\text{m s}^{-1}\)
ⓓ. \(2.0\,\text{m s}^{-1}\)
Correct Answer: \(2.0\,\text{m s}^{-1}\)
Explanation: \( \textbf{Height gained:} \) \(h=0.20\,\text{m}\).
Take \(g=10\,\text{m s}^{-2}\).
At the lowest point, the bob has kinetic energy.
At the highest point it just reaches, its speed is zero.
Energy conservation gives
\[
\frac{1}{2}mv^2=mgh
\]
Cancel \(m\):
\[
\frac{1}{2}v^2=gh
\]
\[
v^2=2gh
\]
Substitute:
\[
v^2=2(10)(0.20)
\]
\[
v^2=4
\]
\[
v=2.0\,\text{m s}^{-1}
\]
Only the vertical height rise matters, not the arc length of the pendulum path.
\( \textbf{Final answer:} \) The required speed is \(2.0\,\text{m s}^{-1}\).
220. Read the case below and answer the question.
A bead slides without friction from point P to point Q along a curved wire. Point Q is \(1.25\,\text{m}\) lower than point P. The bead starts from rest at P, and air resistance is neglected. Take \(g=10\,\text{m s}^{-2}\).
The speed of the bead at Q is
ⓐ. \(2.5\,\text{m s}^{-1}\)
ⓑ. \(5.0\,\text{m s}^{-1}\)
ⓒ. \(12.5\,\text{m s}^{-1}\)
ⓓ. \(25.0\,\text{m s}^{-1}\)
Correct Answer: \(5.0\,\text{m s}^{-1}\)
Explanation: \( \textbf{Height drop:} \) \(h=1.25\,\text{m}\).
The bead starts from rest, so \(K_i=0\).
The wire is frictionless, so mechanical energy is conserved.
The loss of gravitational potential energy becomes kinetic energy:
\[
mgh=\frac{1}{2}mv^2
\]
Cancel \(m\):
\[
v^2=2gh
\]
Substitute:
\[
v^2=2(10)(1.25)
\]
\[
v^2=25
\]
\[
v=5.0\,\text{m s}^{-1}
\]
The curved shape of the wire does not change the speed because gravity's work depends only on vertical height difference.
\( \textbf{Final answer:} \) The speed at Q is \(5.0\,\text{m s}^{-1}\).
