101. A particle of mass \(m\) initially moves with velocity \(v\hat i\). In Case I, its velocity is changed by adding \(v\hat i\). In Case II, starting again from the same initial state, its velocity is changed by adding \(v\hat j\). If the corresponding kinetic-energy increases are \(\Delta K_1\) and \(\Delta K_2\), then
ⓐ. \(\Delta K_1=\Delta K_2\)
ⓑ. \(\Delta K_1=2\Delta K_2\)
ⓒ. \(\Delta K_1=3\Delta K_2\)
ⓓ. \(\Delta K_2=3\Delta K_1\)
Correct Answer: \(\Delta K_1=3\Delta K_2\)
Explanation: The same magnitude of velocity change does not guarantee the same kinetic-energy change because the direction of the added velocity matters.
The initial kinetic energy is
\[
K_i=\frac12mv^2
\]
\(\textbf{Case I — parallel addition:}\)
The final velocity is
\[
\vec v_1=2v\hat i
\]
Hence
\[
K_1=\frac12m(2v)^2=4K_i
\]
Therefore
\[
\Delta K_1=4K_i-K_i=3K_i
\]
\(\textbf{Case II — perpendicular addition:}\)
The final velocity is
\[
\vec v_2=v\hat i+v\hat j
\]
Its squared speed is
\[
v_2^2=v^2+v^2=2v^2
\]
Thus
\[
K_2=2K_i,
\]
and
\[
\Delta K_2=K_i
\]
Therefore
\[
\frac{\Delta K_1}{\Delta K_2}=3
\]
\(\textbf{Final answer:}\)
\[
\boxed{\Delta K_1=3\Delta K_2}
\]
102. Two particles \(A\) and \(B\) initially have equal kinetic energies. Their masses are \(m\) and \(9m\), respectively. Particle \(A\)'s speed is doubled, while particle \(B\)'s speed is increased by \(50\%\). The ratio of their increases in kinetic energy, \(\Delta K_A:\Delta K_B\), is
ⓐ. \(3:2\)
ⓑ. \(12:5\)
ⓒ. \(9:4\)
ⓓ. \(5:3\)
Correct Answer: \(12:5\)
Explanation: Because the particles start with equal kinetic energies, the changes can be compared using fractional energy factors without first finding their actual speeds.
Let each initial kinetic energy be \(K\).
\(\textbf{Particle A:}\)
Doubling the speed multiplies kinetic energy by
\[
2^2=4
\]
Therefore
\[
\Delta K_A=4K-K=3K
\]
\(\textbf{Particle B:}\)
Increasing speed by \(50\%\) multiplies it by \(1.5\).
Hence
\[
K'_B=(1.5)^2K=\frac94K
\]
Therefore
\[
\Delta K_B=\frac94K-K=\frac54K
\]
Thus
\[
\Delta K_A:\Delta K_B
=
3K:\frac54K
\]
\[
\Delta K_A:\Delta K_B=12:5
\]
The different masses do not alter this comparison because equality of the initial kinetic energies has already absorbed the mass-speed difference.
\(\textbf{Final answer:}\)
\[
\boxed{12:5}
\]
103. A particle of fixed mass changes speed successively from \(v_1\) to \(v_2\) and then from \(v_2\) to \(v_3\). The kinetic-energy increases in the two stages are equal and positive. Which relation must hold?
ⓐ. \(v_2-v_1=v_3-v_2\)
ⓑ. \(v_1v_3=v_2^2\)
ⓒ. \(v_1^2+v_3^2=2v_2^2\)
ⓓ. \(v_1+v_3=2v_2\)
Correct Answer: \(v_1^2+v_3^2=2v_2^2\)
Explanation: Equal kinetic-energy increments correspond to equal increments in the square of speed, not necessarily equal increments in speed.
The first increase is
\[
\Delta K_1=\frac12m(v_2^2-v_1^2)
\]
The second is
\[
\Delta K_2=\frac12m(v_3^2-v_2^2)
\]
Given
\[
\Delta K_1=\Delta K_2,
\]
we have
\[
v_2^2-v_1^2=v_3^2-v_2^2
\]
Rearranging,
\[
2v_2^2=v_1^2+v_3^2
\]
Thus the squared speeds, rather than the speeds themselves, form an arithmetic progression.
Equal kinetic-energy gains therefore do not usually correspond to equal speed gains.
The relation required is
\[
\boxed{v_1^2+v_3^2=2v_2^2}
\]
104. Match each change in List I with the corresponding kinetic-energy factor \(K_f/K_i\) in List II.
| List I | Change | List II | \(K_f/K_i\) |
|---|
| P | Mass doubled, speed unchanged | 1 | \(1\) |
| Q | Mass unchanged, speed halved | 2 | \(2\) |
| R | Mass halved, speed doubled | 3 | \(1/4\) |
| S | Velocity reversed with unchanged speed and mass | 4 | \(4\) |
Choose the correct matching.
ⓐ. P-2, Q-3, R-2, S-1
ⓑ. P-2, Q-3, R-4, S-1
ⓒ. P-4, Q-3, R-2, S-1
ⓓ. P-2, Q-1, R-4, S-3
Correct Answer: P-2, Q-3, R-2, S-1
Explanation: Kinetic energy scales as
\[
K\propto mv^2.
\]
For P, doubling mass at fixed speed gives
\[
\frac{K_f}{K_i}=2.
\]
Thus P matches 2.
For Q, halving speed gives
\[
\frac{K_f}{K_i}=\left(\frac12\right)^2=\frac14,
\]
so Q matches 3.
For R, halving mass while doubling speed gives
\[
\frac{K_f}{K_i}
=
\frac12(2)^2=2,
\]
so R also matches 2.
For S, reversing velocity leaves speed unchanged:
\[
\frac{K_f}{K_i}=1.
\]
Thus S matches 1.
The repeated match to factor \(2\) is physically valid because two different changes can produce the same final energy factor.
Therefore the matching is P-2, Q-3, R-2, S-1.
105. A particle moves along the positive \(x\)-direction in frame \(S\). Its kinetic energy is \(25\,\text{J}\) in \(S\) and \(9\,\text{J}\) in a frame moving at \(+2\,\text{m s}^{-1}\) relative to \(S\). The particle moves faster than \(2\,\text{m s}^{-1}\) in \(S\). What are its mass, its speed in \(S\), and its kinetic energy in a frame moving at \(+7\,\text{m s}^{-1}\) relative to \(S\)?
ⓐ. \(4\,\text{kg},\;5/\sqrt2\,\text{m s}^{-1},\;9\,\text{J}\)
ⓑ. \(1\,\text{kg},\;5\sqrt2\,\text{m s}^{-1},\;2\,\text{J}\)
ⓒ. \(2\,\text{kg},\;5\,\text{m s}^{-1},\;16\,\text{J}\)
ⓓ. \(2\,\text{kg},\;5\,\text{m s}^{-1},\;4\,\text{J}\)
Correct Answer: \(2\,\text{kg},\;5\,\text{m s}^{-1},\;4\,\text{J}\)
Explanation: Let the particle speed in \(S\) be \(v\).
\(\textbf{Energy in frame }S\textbf{:}\)
\[
\frac12mv^2=25
\]
In the frame moving at \(+2\,\text{m s}^{-1}\), the particle speed is \(v-2\) because \(v\gt 2\).
Thus
\[
\frac12m(v-2)^2=9
\]
\(\textbf{Use the ratio of the two measurements:}\)
\[
\frac{(v-2)^2}{v^2}=\frac9{25}
\]
Taking the positive square root,
\[
\frac{v-2}{v}=\frac35
\]
Hence
\[
5v-10=3v,
\]
so
\[
v=5\,\text{m s}^{-1}
\]
Substitution into the first energy equation gives
\[
25=\frac12m(25),
\]
therefore
\[
m=2\,\text{kg}
\]
\(\textbf{Frame moving at }+7\,\text{m s}^{-1}\textbf{:}\)
The relative velocity is
\[
v'=5-7=-2\,\text{m s}^{-1}
\]
Hence
\[
K'=\frac12(2)(-2)^2=4\,\text{J}
\]
\(\textbf{Final answer:}\)
\[
\boxed{m=2\,\text{kg},\quad v=5\,\text{m s}^{-1},\quad K'=4\,\text{J}}
\]
106. A particle of fixed mass has kinetic energy \(K\) at speed \(v\). Its speed is changed first to \(\sqrt3\,v\) and then to \(2v\). Which statement correctly compares the two successive kinetic-energy increases?
ⓐ. The two increases are equal
ⓑ. The second increase is twice the first
ⓒ. The first increase is twice the second
ⓓ. The second increase is three times the first
Correct Answer: The first increase is twice the second
Explanation: The kinetic energies at the three speeds are
\[
K_1=K,
\]
\[
K_2=3K,
\]
and
\[
K_3=4K.
\]
Therefore the first increase is
\[
\Delta K_1=3K-K=2K.
\]
The second increase is
\[
\Delta K_2=4K-3K=K.
\]
Hence
\[
\Delta K_1=2\Delta K_2.
\]
The second speed interval, from \(\sqrt3v\) to \(2v\), is not the larger energy interval merely because it occurs at the higher speeds.
The decisive quantities are the differences of the squared speeds.
Thus the first kinetic-energy increase is twice the second.
107. A particle of mass \(m\) has kinetic energy \(K\). Another particle has mass \(4m\) and kinetic energy \(9K\). Both particles then have their speeds reduced by the same fraction \(1/3\). If their kinetic-energy losses are \(L_1\) and \(L_2\), respectively, then
ⓐ. \(L_1:L_2=1:3\)
ⓑ. \(L_1:L_2=1:4\)
ⓒ. \(L_1:L_2=1:9\)
ⓓ. \(L_1:L_2=4:9\)
Correct Answer: \(L_1:L_2=1:9\)
Explanation: Reducing speed by one-third means the final speed is
\[
\frac23
\]
of the initial speed.
For either particle, the final kinetic energy is therefore
\[
K_f=\left(\frac23\right)^2K_i=\frac49K_i
\]
Hence the kinetic-energy loss is
\[
L=K_i-K_f
\]
\[
L=K_i-\frac49K_i=\frac59K_i
\]
Thus each particle loses the same fraction \(5/9\) of its own initial kinetic energy.
The initial energies are \(K\) and \(9K\), so
\[
L_1=\frac59K,
\]
while
\[
L_2=\frac59(9K)=5K
\]
Therefore
\[
\boxed{L_1:L_2=1:9}
\]
The different masses are already reflected in the stated initial kinetic energies and need not be introduced again.
108. A \(3\,\text{kg}\) particle changes speed from \(4\,\text{m s}^{-1}\) to \(v\), and its kinetic energy decreases by \(18\,\text{J}\). A \(1\,\text{kg}\) particle changes speed from \(6\,\text{m s}^{-1}\) to \(u\), and its kinetic energy increases by the same magnitude \(18\,\text{J}\). The pair \((v,u)\) is
ⓐ. \((2\,\text{m s}^{-1},\,6\sqrt2\,\text{m s}^{-1})\)
ⓑ. \((\sqrt{12}\,\text{m s}^{-1},\,6\sqrt2\,\text{m s}^{-1})\)
ⓒ. \((2\,\text{m s}^{-1},\,6\sqrt3\,\text{m s}^{-1})\)
ⓓ. \((\sqrt{12}\,\text{m s}^{-1},\,6\sqrt3\,\text{m s}^{-1})\)
Correct Answer: \((2\,\text{m s}^{-1},\,6\sqrt2\,\text{m s}^{-1})\)
Explanation: \(\textbf{First particle — energy decrease:}\)
Its initial kinetic energy is
\[
K_{1i}=\frac12(3)(4^2)=24\,\text{J}
\]
A decrease of \(18\,\text{J}\) gives
\[
K_{1f}=24-18=6\,\text{J}
\]
Therefore
\[
6=\frac12(3)v^2
\]
Hence
\[
v^2=4,
\qquad
v=2\,\text{m s}^{-1}
\]
\(\textbf{Second particle — energy increase:}\)
Its initial kinetic energy is
\[
K_{2i}=\frac12(1)(6^2)=18\,\text{J}
\]
After an increase of \(18\,\text{J}\),
\[
K_{2f}=36\,\text{J}
\]
Therefore
\[
36=\frac12u^2
\]
Thus
\[
u^2=72,
\]
so
\[
u=6\sqrt2\,\text{m s}^{-1}
\]
The correct pair is therefore
\[
\boxed{(v,u)=(2\,\text{m s}^{-1},\,6\sqrt2\,\text{m s}^{-1})}
\]
Hence the correct option is A.
109. Consider the following statements about kinetic energy.
Statement I: A negative change in kinetic energy means the final kinetic energy is negative.
Statement II: Equal changes in speed produce equal changes in kinetic energy for particles of equal mass.
Statement III: Equal kinetic energies for particles of unequal masses require their speeds to vary inversely as the square root of mass.
Statement IV: A change in velocity direction at unchanged speed leaves kinetic energy unchanged.
Which of the following is correct?
ⓐ. Statements I and II only
ⓑ. Statements II and III only
ⓒ. Statements III and IV only
ⓓ. Statements I, III and IV only
Correct Answer: Statements III and IV only
Explanation: Statement I is incorrect because
\[
\Delta K=K_f-K_i\lt 0
\]
only means that the final kinetic energy is smaller than the initial kinetic energy.
It does not imply \(K_f\lt 0\).
Classical translational kinetic energy remains non-negative.
Statement II is also incorrect.
For equal masses,
\[
\Delta K=\frac12m(v_f^2-v_i^2),
\]
so the same numerical speed change produces different energy changes when the initial speeds differ.
Statement III is correct.
For equal kinetic energies,
\[
\frac12mv^2=\text{constant},
\]
which gives
\[
v\propto\frac1{\sqrt m}.
\]
Statement IV is correct because kinetic energy depends on speed, not on velocity direction.
Therefore only Statements III and IV are correct.
110. Three particles \(A\), \(B\) and \(C\) initially have equal kinetic energies. Their masses are \(m\), \(4m\) and \(9m\), respectively. Each particle's speed is then increased by the same absolute amount equal to the initial speed of particle \(C\). If their final kinetic energies are \(K_A',K_B',K_C'\), the ratio \(K_A':K_B':K_C'\) is
ⓐ. \(16:16:16\)
ⓑ. \(4:9:16\)
ⓒ. \(25:16:9\)
ⓓ. \(16:25:36\)
Correct Answer: \(16:25:36\)
Explanation: The initial equal-energy condition first determines the speed ratios.
Let the initial speed of particle \(C\) be \(u\).
Because
\[
\frac12mv_A^2
=
\frac12(4m)v_B^2
=
\frac12(9m)u^2,
\]
the speeds vary inversely as the square roots of the masses.
Thus
\[
v_A=3u,
\qquad
v_B=\frac32u,
\qquad
v_C=u
\]
Each particle now receives the same absolute speed increment \(u\).
Therefore
\[
v_A'=4u,
\qquad
v_B'=\frac52u,
\qquad
v_C'=2u
\]
\(\textbf{Final kinetic-energy factors:}\)
For \(A\),
\[
K_A'\propto m(4u)^2=16mu^2
\]
For \(B\),
\[
K_B'\propto4m\left(\frac52u\right)^2
\]
\[
K_B'\propto25mu^2
\]
For \(C\),
\[
K_C'\propto9m(2u)^2=36mu^2
\]
Therefore
\[
K_A':K_B':K_C'=16:25:36
\]
Although all three began with equal kinetic energies and received the same absolute speed increase, the final energy gains differ because their initial speeds and masses differ.
\(\textbf{Final answer:}\)
\[
\boxed{16:25:36}
\]
111. Consider the following statements about potential energy.
Statement I: Potential energy is associated with the configuration of a system under a conservative interaction.
Statement II: The zero level of potential energy may be chosen conveniently without changing physically measurable potential-energy differences.
Statement III: Potential energy of a system can never be negative.
Statement IV: If the potential energy is zero at a configuration, the conservative force must be zero there.
Which of the following is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements II and IV only
ⓒ. Statements I and II only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I and II only
Explanation: Statement I is correct because potential energy describes an energy associated with the configuration of an interacting system rather than simply the motion of one particle.
For a conservative interaction, differences in potential energy are physically related to work done by the conservative force.
Statement II is also correct.
An arbitrary constant may be added to every potential-energy value without changing any potential-energy difference.
Consequently, physically relevant predictions based on \(\Delta U\) remain unchanged.
Statement III is incorrect because the numerical sign of an absolute potential-energy value depends on the chosen zero.
A configuration may therefore be assigned negative, zero or positive potential energy under different valid reference choices.
Statement IV is incorrect because choosing \(U=0\) at a configuration does not imply that the conservative interaction or force vanishes there.
The zero of \(U\) is a reference convention, not a condition specifying zero force.
112. A \(2\,\text{kg}\) body moves near Earth's surface where \(g=10\,\text{m s}^{-2}\). Three configurations have heights \(y_A=1\,\text{m}\), \(y_B=4\,\text{m}\) and \(y_C=-2\,\text{m}\). In Convention I, \(U=0\) at \(y=0\). In Convention II, \(U=0\) at \(y=3\,\text{m}\).
| Configuration | \(U\) in Convention I | \(U\) in Convention II |
|---|
| A | \(20\,\text{J}\) | \(-40\,\text{J}\) |
| B | \(80\,\text{J}\) | \(?\) |
| C | \(-40\,\text{J}\) | \(-100\,\text{J}\) |
What are the missing potential energy at \(B\) in Convention II and the work done by gravity when the body moves from \(B\) to \(C\)?
ⓐ. \(20\,\text{J},\;-120\,\text{J}\)
ⓑ. \(80\,\text{J},\;+120\,\text{J}\)
ⓒ. \(-20\,\text{J},\;+60\,\text{J}\)
ⓓ. \(20\,\text{J},\;+120\,\text{J}\)
Correct Answer: \(20\,\text{J},\;+120\,\text{J}\)
Explanation: Changing the zero from \(y=0\) to \(y=3\,\text{m}\) subtracts the same gravitational reference contribution \(mg(3)\) from every potential-energy value for this body.
\(\textbf{Reference shift:}\)
\[
mg(3)=(2)(10)(3)=60\,\text{J}
\]
Therefore at \(B\),
\[
U_B'=80-60=20\,\text{J}
\]
\(\textbf{Potential-energy change from }B\textbf{ to }C\textbf{:}\)
Using Convention II,
\[
\Delta U=U_C'-U_B'
\]
\[
\Delta U=-100-20=-120\,\text{J}
\]
Gravity is conservative, so
\[
W_g=-\Delta U
\]
Hence
\[
W_g=+120\,\text{J}
\]
The same \(+120\,\text{J}\) would be obtained using Convention I because changing the zero does not change \(\Delta U\).
\(\textbf{Final answer:}\)
\[
\boxed{U_B'=20\,\text{J},\qquad W_g=120\,\text{J}}
\]
113.
A \(2\,\text{kg}\) body moves from point \(P\), \(1\,\text{m}\) above a reference plane, to point \(Q\), \(6\,\text{m}\) above the same plane. Along Route I it moves directly from \(P\) to \(Q\). Along Route II it first rises to a point \(10\,\text{m}\) above the reference plane and then descends to \(Q\). Take \(g=10\,\text{m s}^{-2}\).
Which statement correctly compares the gravitational potential-energy change and gravitational work for the two routes?
ⓐ. Route II has zero gravitational work because its upward and downward portions cancel completely
ⓑ. Route II has a larger positive \(\Delta U\) because the body rises to \(10\,\text{m}\)
ⓒ. For both routes, \(\Delta U=-100\,\text{J}\) and \(W_g=+100\,\text{J}\)
ⓓ. For both routes, \(\Delta U=+100\,\text{J}\) and \(W_g=-100\,\text{J}\)
Correct Answer: For both routes, \(\Delta U=+100\,\text{J}\) and \(W_g=-100\,\text{J}\)
Explanation: Gravitational potential-energy change near Earth depends only on the initial and final heights.
The intermediate maximum height on Route II does not alter the overall change between \(P\) and \(Q\).
\(\textbf{Endpoint height change:}\)
\[
\Delta y=6-1=5\,\text{m}
\]
Therefore
\[
\Delta U=mg\Delta y.
\]
\[
\Delta U=(2)(10)(5)=100\,\text{J}
\]
Gravity is conservative, so
\[
W_g=-\Delta U=-100\,\text{J}
\]
On Route II, gravity does negative work during the larger upward part and positive work during the later descent.
Those contributions combine to the same \(-100\,\text{J}\) obtained on Route I.
Thus travelled distance and intermediate height do not determine the net gravitational potential-energy change when the endpoints are fixed.
114. A conservative force does \(+36\,\text{J}\) of work when a system changes from configuration \(A\) to configuration \(B\). In one potential-energy convention, \(U_A=-10\,\text{J}\). The zero of potential energy is then changed so that every potential-energy value increases by \(50\,\text{J}\). What are the new value of \(U_B\) and the work done by the same conservative force from \(B\) back to \(A\)?
ⓐ. \(4\,\text{J},\;+36\,\text{J}\)
ⓑ. \(4\,\text{J},\;-36\,\text{J}\)
ⓒ. \(76\,\text{J},\;-36\,\text{J}\)
ⓓ. \(-46\,\text{J},\;+36\,\text{J}\)
Correct Answer: \(4\,\text{J},\;-36\,\text{J}\)
Explanation: \(\textbf{Use conservative work to find the potential change:}\)
\[
W_c=-\Delta U
\]
Therefore
\[
+36=-(U_B-U_A)
\]
Hence
\[
U_B-U_A=-36\,\text{J}
\]
With \(U_A=-10\,\text{J}\),
\[
U_B=-10-36=-46\,\text{J}
\]
\(\textbf{Apply the new zero convention:}\)
Every potential-energy value increases by \(50\,\text{J}\), so
\[
U_B'=-46+50=4\,\text{J}
\]
The shift does not alter the potential-energy difference.
\(\textbf{Reverse the motion:}\)
Reversing the endpoints reverses the conservative work:
\[
W_{B\to A}=-36\,\text{J}
\]
\(\textbf{Final answer:}\)
\[
\boxed{U_B'=4\,\text{J},\qquad W_{B\to A}=-36\,\text{J}}
\]
115. The potential energies of three configurations of the same conservative system are shown under two different reference conventions.
| Configuration | Convention I | Convention II |
|---|
| A | \(12\,\text{J}\) | \(42\,\text{J}\) |
| B | \(-8\,\text{J}\) | \(22\,\text{J}\) |
| C | \(20\,\text{J}\) | \(?\) |
Which pair gives the missing value in Convention II and the work done by the conservative force from \(A\) to \(C\)?
ⓐ. \(30\,\text{J},\;+8\,\text{J}\)
ⓑ. \(50\,\text{J},\;+8\,\text{J}\)
ⓒ. \(50\,\text{J},\;-8\,\text{J}\)
ⓓ. \(20\,\text{J},\;-28\,\text{J}\)
Correct Answer: \(50\,\text{J},\;-8\,\text{J}\)
Explanation: The difference between the two conventions is the same additive constant for every configuration.
\(\textbf{Determine the reference shift:}\)
From configuration \(A\),
\[
42-12=30\,\text{J}
\]
The same shift is confirmed by \(B\):
\[
22-(-8)=30\,\text{J}
\]
Therefore
\[
U_C'=20+30=50\,\text{J}
\]
\(\textbf{Conservative work from }A\textbf{ to }C\textbf{:}\)
\[
W_c=-(U_C-U_A)
\]
Using Convention I,
\[
W_c=-(20-12)=-8\,\text{J}
\]
Using Convention II would give
\[
W_c=-(50-42)=-8\,\text{J}
\]
again.
Thus an additive change in potential-energy zero alters absolute values but not conservative work between fixed configurations.
\(\textbf{Final answer:}\)
\[
\boxed{U_C'=50\,\text{J},\qquad W_{A\to C}=-8\,\text{J}}
\]
116. A particle moves under an unchanged conservative interaction through the sequence \(A\to B\to C\). The work done by the conservative force is \(+20\,\text{J}\) from \(A\) to \(B\) and \(-35\,\text{J}\) from \(B\) to \(C\). It then returns from \(C\) to \(A\) along any other path. What are the work done by the conservative force from \(C\) to \(A\) and its total work over the complete closed motion?
ⓐ. \(-55\,\text{J},\;-70\,\text{J}\)
ⓑ. \(-15\,\text{J},\;0\)
ⓒ. \(+55\,\text{J},\;+40\,\text{J}\)
ⓓ. \(+15\,\text{J},\;0\)
Correct Answer: \(+15\,\text{J},\;0\)
Explanation: \(\textbf{Combine the first two legs:}\)
\[
W_{A\to C}=W_{A\to B}+W_{B\to C}
\]
\[
W_{A\to C}=20-35=-15\,\text{J}
\]
For a conservative interaction, the work between two configurations depends only on the endpoints.
Therefore reversing the endpoints changes the sign:
\[
W_{C\to A}=+15\,\text{J}
\]
The path used for the return does not alter this value.
\(\textbf{Closed-motion check:}\)
\[
W_{\text{closed}}
=
W_{A\to C}+W_{C\to A}
\]
\[
W_{\text{closed}}=-15+15=0
\]
Equivalently, the system returns to its original potential energy, so the total \(\Delta U\) is zero.
\(\textbf{Final answer:}\)
\[
\boxed{W_{C\to A}=15\,\text{J},\qquad W_{\text{closed}}=0}
\]
117. Assertion: A body may have negative gravitational potential energy relative to a chosen zero level, yet gravity may do either positive or negative work during a subsequent displacement.
Reason: Gravitational work is determined by \(W_g=-\Delta U\), so its sign depends on the change in potential energy rather than on the absolute sign of \(U\).
ⓐ. Both Assertion and Reason are correct, and the Reason correctly explains the Assertion
ⓑ. Both Assertion and Reason are correct, but the Reason does not correctly explain the Assertion
ⓒ. Assertion is correct, but Reason is incorrect
ⓓ. Assertion is incorrect, but Reason is correct
Correct Answer: Both Assertion and Reason are correct, and the Reason correctly explains the Assertion
Explanation: The Assertion is correct because an absolute gravitational potential-energy value depends on the chosen reference level.
Choosing a zero above the body can make its numerical \(U\) negative without changing the physical gravitational force.
The work done by gravity during a motion is governed by the difference between final and initial potential energies.
For an upward displacement near Earth,
\[
\Delta U\gt 0
\]
and therefore
\[
W_g\lt 0.
\]
For a downward displacement,
\[
\Delta U\lt 0
\]
and therefore
\[
W_g\gt 0.
\]
These statements remain true whether the initial and final absolute \(U\) values are positive, negative or zero.
Thus the Reason correctly explains why the absolute sign of potential energy does not determine the sign of gravitational work.
118. A \(3\,\text{kg}\) body moves near Earth's surface from height \(y_i=5\,\text{m}\) to \(y_f=-2\,\text{m}\). The chosen zero of gravitational potential energy is the horizontal plane \(y=4\,\text{m}\). Take \(g=10\,\text{m s}^{-2}\). What are the final potential energy and the work done by gravity during the motion?
ⓐ. \(-60\,\text{J},\;+210\,\text{J}\)
ⓑ. \(-180\,\text{J},\;-210\,\text{J}\)
ⓒ. \(+180\,\text{J},\;+210\,\text{J}\)
ⓓ. \(-180\,\text{J},\;+210\,\text{J}\)
Correct Answer: \(-180\,\text{J},\;+210\,\text{J}\)
Explanation: With zero potential energy at \(y=4\,\text{m}\),
\[
U=mg(y-4)
\]
\(\textbf{Initial potential energy:}\)
\[
U_i=(3)(10)(5-4)=30\,\text{J}
\]
\(\textbf{Final potential energy:}\)
\[
U_f=(3)(10)(-2-4)
\]
\[
U_f=-180\,\text{J}
\]
Therefore
\[
\Delta U=U_f-U_i
\]
\[
\Delta U=-180-30=-210\,\text{J}
\]
Gravity does the negative of this potential-energy change:
\[
W_g=-\Delta U=+210\,\text{J}
\]
The negative final value of \(U\) merely reflects that the final position lies below the selected zero plane.
\(\textbf{Final answer:}\)
\[
\boxed{U_f=-180\,\text{J},\qquad W_g=210\,\text{J}}
\]
119.
A \(4\,\text{kg}\) body moves from a point \(2\,\text{m}\) above ground to a point \(7\,\text{m}\) above ground. Along Route X it goes directly between the two points. Along Route Y it first rises from \(2\,\text{m}\) to \(12\,\text{m}\) and then descends from \(12\,\text{m}\) to \(7\,\text{m}\). Take \(g=10\,\text{m s}^{-2}\).
Which option gives the gravitational work along Route X and the gravitational works on the two successive portions of Route Y?
ⓐ. \(-200\,\text{J};\;-200\,\text{J},0\)
ⓑ. \(-200\,\text{J};\;-400\,\text{J},+200\,\text{J}\)
ⓒ. \(+200\,\text{J};\;-400\,\text{J},+600\,\text{J}\)
ⓓ. \(-400\,\text{J};\;-400\,\text{J},+200\,\text{J}\)
Correct Answer: \(-200\,\text{J};\;-400\,\text{J},+200\,\text{J}\)
Explanation: \(\textbf{Route X — endpoint calculation:}\)
The net rise is
\[
\Delta y=7-2=5\,\text{m}
\]
Hence
\[
W_{g,X}=-mg\Delta y
\]
\[
W_{g,X}=-(4)(10)(5)=-200\,\text{J}
\]
\(\textbf{Route Y — upward portion:}\)
The body rises from \(2\,\text{m}\) to \(12\,\text{m}\), a rise of \(10\,\text{m}\):
\[
W_{g,1}=-(4)(10)(10)=-400\,\text{J}
\]
\(\textbf{Route Y — downward portion:}\)
The body then descends through \(5\,\text{m}\):
\[
W_{g,2}=+(4)(10)(5)=+200\,\text{J}
\]
Thus
\[
W_{g,Y}=-400+200=-200\,\text{J}
\]
The two routes have different intermediate motions but the same endpoints, so their total gravitational works are equal.
\(\textbf{Final answer:}\)
\[
\boxed{-200\,\text{J};\;-400\,\text{J},+200\,\text{J}}
\]
120. Body \(P\) of mass \(2m\) moves vertically from height \(h\) to \(4h\). Body \(Q\) of mass \(3m\) moves from height \(-2h\) to height \(0\). Assume constant \(g\). The zero level of gravitational potential energy is then shifted from \(y=0\) to \(y=h\) for both bodies. Which statement is correct?
ⓐ. Initially \(\Delta U_P=2\Delta U_Q\), but the shift makes them equal
ⓑ. Initially \(\Delta U_Q=2\Delta U_P\), and both changes decrease after the shift
ⓒ. \(\Delta U_P=\Delta U_Q=6mgh\), and both changes are unchanged by shifting the zero level
ⓓ. \(\Delta U_P=\Delta U_Q=3mgh\), but both become zero after the shift
Correct Answer: \(\Delta U_P=\Delta U_Q=6mgh\), and both changes are unchanged by shifting the zero level
Explanation: \(\textbf{Body P:}\)
Its vertical displacement is
\[
\Delta y_P=4h-h=3h.
\]
Therefore
\[
\Delta U_P=(2m)g(3h)=6mgh
\]
\(\textbf{Body Q:}\)
Its vertical displacement is
\[
\Delta y_Q=0-(-2h)=2h
\]
Hence
\[
\Delta U_Q=(3m)g(2h)=6mgh
\]
Thus the two potential-energy increases are equal despite the different masses and height changes.
Changing the zero from \(y=0\) to \(y=h\) shifts every potential-energy value for each individual body by a constant.
Such a shift changes neither body's final-minus-initial difference.
Therefore
\[
\Delta U_P=\Delta U_Q=6mgh
\]
before and after the reference change.
\(\textbf{Final answer:}\)
\[
\Delta U_P=\Delta U_Q=6mgh.
\]
