Work and Energy — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Work and Energy MCQs with step-by-step solutions (33 questions). Part of Work, Energy and Power. Practise online on Prepizo — no login needed.
▶ Practise Work and Energy online (free)
Questions with solutions
Q1 — Work and Energy · medium
A particle is released from height $S$ from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively
A. $\dfrac{S}{4},\ \dfrac{3gS}{2}$
B. $\dfrac{S}{4},\ \dfrac{\sqrt{3gS}}{2}$
C. $\dfrac{S}{2},\ \dfrac{\sqrt{3gS}}{2}$
D. $\dfrac{S}{4},\ \sqrt{\dfrac{3gS}{2}}$ ✓ Correct
Solution: At height $x$, KE $=3$(PE) $\Rightarrow$ KE $=3mgx$. Energy is conserved, so $mgS=mgx+3mgx=4mgx\Rightarrow x=\dfrac{S}{4}$. Also $\dfrac{1}{2}mv^2=3mgx=\dfrac{3}{4}mgS\Rightarrow v=\sqrt{\dfrac{3gS}{2}}$.
Q2 — Work and Energy · medium
The energy required to break one bond in DNA is $10^{-20}$ J. This value (in eV) is nearly
A. 0.6
B. 0.06 ✓ Correct
C. 0.006
D. 6
Solution: $E=\dfrac{10^{-20}}{1.6\times10^{-19}}$ eV $=0.06$ eV.
Q3 — Work and Energy · medium
A force $F=20+10y$ acts on a particle in $y$-direction, where $F$ is in newton and $y$ in meter. Work done by this force to move the particle from $y=0$ to $y=1$ m is
A. 5 J
B. 25 J ✓ Correct
C. 20 J
D. 30 J
Solution: $W=\int_0^1 (20+10y)\,dy=\left[20y+\dfrac{10y^2}{2}\right]_0^1=20(1)+5(1)^2=25$ J.
Q4 — Work and Energy · medium
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to $8\times10^{-4}$ J by the end of the second revolution after the beginning of the motion?
A. 0.15 m/s$^2$
B. 0.18 m/s$^2$
C. 0.2 m/s$^2$
D. 0.1 m/s$^2$ ✓ Correct
Solution: $\dfrac{1}{2}mv^2=8\times10^{-4}$ J $\Rightarrow v^2=\dfrac{16\times10^{-4}}{0.01}=16\times10^{-2}$. For 2 revolutions, $v^2=2a_t(4\pi r)\Rightarrow a_t=\dfrac{v^2}{8\pi r}=\dfrac{16\times10^{-2}}{8\times3.14\times6.4\times10^{-2}}=0.1$ m/s$^2$.
Q5 — Work and Energy · medium
A particle moves from a point $(-2\hat{i}+5\hat{j})$ to $(4\hat{j}+3\hat{k})$ when a force of $(4\hat{i}+3\hat{j})$ N is applied. How much work has been done by the force?
A. 8 J
B. 11 J
C. 5 J ✓ Correct
D. 2 J
Solution: Displacement $\Delta \mathbf{s}=\mathbf{r_2}-\mathbf{r_1}=(4\hat{j}+3\hat{k})-(-2\hat{i}+5\hat{j})=2\hat{i}-\hat{j}+3\hat{k}$. $W=\mathbf{F}\cdot\Delta \mathbf{s}=(4\hat{i}+3\hat{j})\cdot(2\hat{i}-\hat{j}+3\hat{k})=8-3=5$ J.
Q6 — Work and Energy · medium
Two similar springs $P$ and $Q$ have spring constants $K_P$ and $K_Q$, such that $K_P>K_Q$. They are stretched, first by the same amount (case $a$), then by the same force (case $b$). The work done by the springs $W_P$ and $W_Q$ are related as, in case ($a$) and case ($b$), respectively
A. $W_P=W_Q\ ;\ W_P>W_Q$
B. $W_P=W_Q\ ;\ W_P=W_Q$
C. $W_P>W_Q\ ;\ W_Q>W_P$ ✓ Correct
D. $W_P<W_Q\ ;\ W_Q<W_P$
Solution: Same extension: $W=\dfrac{1}{2}Kx^2$, so $\dfrac{W_P}{W_Q}=\dfrac{K_P}{K_Q}>1\Rightarrow W_P>W_Q$. Same force: $W=\dfrac{1}{2}\dfrac{F^2}{K}$, so $\dfrac{W_P}{W_Q}=\dfrac{K_Q}{K_P}<1\Rightarrow W_P<W_Q$, i.e. $W_Q>W_P$.
Q7 — Work and Energy · medium
A block of mass 10 kg, moving in $x$-direction with a constant speed of 10 ms$^{-1}$, is subjected to a retarding force $F=0.1x$ J/m during its travel from $x=20$ m to 30 m. Its final KE will be
A. 475 J ✓ Correct
B. 450 J
C. 275 J
D. 250 J
Solution: By work-energy theorem, $K_f=K_i+\int_{20}^{30}(-0.1x)\,dx=\dfrac{1}{2}\times10\times10^2-0.05[30^2-20^2]=500-0.05(900-400)=500-25=475$ J.
Q8 — Work and Energy · medium
Two particles of masses $m_1, m_2$ move with initial velocities $u_1$ and $u_2$. On collision, one of the particles get excited to higher level, after absorbing energy $\varepsilon$. If final velocities of particles be $v_1$ and $v_2$, then we must have
A. $m_1^2u_1+m_2^2u_2-\varepsilon=m_1^2v_1+m_2^2v_2$
B. $\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2-\varepsilon$
C. $\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2-\varepsilon=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2$ ✓ Correct
D. $\dfrac{1}{2}m_1^2u_1^2+\dfrac{1}{2}m_2^2u_2^2+\varepsilon=\dfrac{1}{2}m_1^2v_1^2+\dfrac{1}{2}m_2^2v_2^2$
Solution: By conservation of energy, total initial KE $=$ total final KE $+\varepsilon$: $\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2+\varepsilon\Rightarrow \dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2-\varepsilon=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2$.
Q9 — Work and Energy · medium
A mass $m$ moves in a circle on a smooth horizontal plane with velocity $v_0$ at a radius $R_0$. The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally $m$ moves in a circle of radius $\dfrac{R_0}{2}$. The final value of the kinetic energy is
A. $mv_0^2$
B. $\dfrac{1}{4}mv_0^2$
C. $2mv_0^2$ ✓ Correct
D. $\dfrac{1}{2}mv_0^2$
Solution: Conserving angular momentum, $mv_0R_0=mv'\dfrac{R_0}{2}\Rightarrow v'=2v_0$. Final KE $=\dfrac{1}{2}mv'^2=\dfrac{1}{2}m(2v_0)^2=2mv_0^2$.
Q10 — Work and Energy · medium
A ball is thrown vertically downwards from a height of 20 m with an initial velocity $v_0$. It collides with the ground, loses 50% of its energy in collision and rebounds to the same height. The initial velocity $v_0$ is (Take, $g=10$ ms$^{-2}$)
A. 14 ms$^{-1}$
B. 20 ms$^{-1}$ ✓ Correct
C. 28 ms$^{-1}$
D. 10 ms$^{-1}$
Solution: Rebound speed $v=\sqrt{2gh}=\sqrt{2\times10\times20}=20$ m/s, so energy just after rebound $=\dfrac{1}{2}mv^2=200m$. Since 50% is lost in collision, energy just before collision $=400m$. By conservation, $\dfrac{1}{2}mv_0^2+mgh=400m\Rightarrow \dfrac{1}{2}mv_0^2+m\times10\times20=400m\Rightarrow v_0=20$ m/s.
Q11 — Work and Energy · medium
A uniform force of $(3\hat{i}+\hat{j})$ N acts on a particle of mass 2 kg. Hence, the particle is displaced from position $(2\hat{i}+\hat{k})$ m to position $(4\hat{i}+3\hat{j}-\hat{k})$ m. The work done by the force on the particle is
A. 9 J ✓ Correct
B. 6 J
C. 13 J
D. 15 J
Solution: Given $\mathbf{F}=3\hat{i}+\hat{j}$, $\mathbf{r}_1=(2\hat{i}+\hat{k})$ m and $\mathbf{r}_2=(4\hat{i}+3\hat{j}-\hat{k})$ m. Displacement $\mathbf{s}=\mathbf{r}_2-\mathbf{r}_1=(2\hat{i}+3\hat{j}-2\hat{k})$ m. $W=\mathbf{F}\cdot\mathbf{s}=(3\hat{i}+\hat{j})\cdot(2\hat{i}+3\hat{j}-2\hat{k})=3\times2+3+0=6+3=9$ J.
Q12 — Work and Energy · medium
The potential energy of a particle in a force field is $U=\dfrac{A}{r^2}-\dfrac{B}{r}$, where $A$ and $B$ are positive constants and $r$ is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is
A. $B/2A$
B. $2A/B$ ✓ Correct
C. $A/B$
D. $B/A$
Solution: For stable equilibrium, $F=-\dfrac{dU}{dr}=0$. $\dfrac{dU}{dr}=-2Ar^{-3}+Br^{-2}=0\Rightarrow 0=-\dfrac{2A}{r^3}+\dfrac{B}{r^2}\Rightarrow \dfrac{2A}{r}=B\Rightarrow r=\dfrac{2A}{B}$.
Q13 — Work and Energy · medium
The potential energy of a system increases, if work is done
A. by the system against a conservative force ✓ Correct
B. by the system against a non-conservative force
C. upon the system by a conservative force
D. upon the system by a non-conservative force
Solution: The potential energy of a system increases if work is done by the system against a conservative force, since $-\Delta U=W_{\text{conservative force}}$.
Q14 — Work and Energy · medium
Force $F$ on a particle moving in a straight line varies with distance $d$ as shown in the figure. The work done on the particle during its displacement of 12 m is
A. 21 J
B. 26 J
C. 13 J ✓ Correct
D. 18 J
Solution: Work done equals the area under the $F$-$d$ graph. $W=2\times(7-3)+\dfrac{1}{2}\times2\times(12-7)=8+5=13$ J.
Q15 — Work and Energy · medium
A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $k$. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be
A. $Mg/k$
B. $2Mg/k$ ✓ Correct
C. $4Mg/k$
D. $Mg/2k$
Solution: Let $x$ be the extension in the spring. Applying conservation of energy, $Mgx-\dfrac{1}{2}kx^2=0-0\Rightarrow x=\dfrac{2Mg}{k}$.
Q16 — Work and Energy · medium
A body of mass 1 kg is thrown upwards with a velocity $20\ \text{ms}^{-1}$. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (Take $g=10\ \text{ms}^{-2}$)
A. 20 J ✓ Correct
B. 30 J
C. 40 J
D. 10 J
Solution: Loss of energy $=\text{KE}-\text{PE}=\dfrac{1}{2}mv^2-mgh=\dfrac{1}{2}\times1\times400-1\times18\times10=200-180=20$ J.
Q17 — Work and Energy · medium
300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking $g=10\ \text{m/s}^2$, work done against friction is
A. 200 J
B. 100 J ✓ Correct
C. zero
D. 1000 J
Solution: Net work $W=W_g+W_f$, where $W_g=mgh=2\times10\times10=200$ J and $W=300$ J. So $W_f=300-200=100$ J.
Q18 — Work and Energy · medium
A body of mass 3 kg is under a constant force, which causes a displacement $s$ in metre in it, given by the relation $s=\dfrac{1}{3}t^2$, where $t$ is in second. Work done by the force in 2 s is
A. $\dfrac{5}{19}$ J
B. $\dfrac{3}{8}$ J
C. $\dfrac{8}{3}$ J ✓ Correct
D. $\dfrac{19}{5}$ J
Solution: $W=F\times s=mas=m\dfrac{d^2s}{dt^2}s$. With $s=\dfrac{1}{3}t^2$, $\dfrac{d^2s}{dt^2}=\dfrac{2}{3}$. So $W=\dfrac{2}{3}ms=\dfrac{2}{3}m\times\dfrac{1}{3}t^2=\dfrac{2}{9}mt^2$. For $m=3$ kg, $t=2$ s, $W=\dfrac{2}{9}\times3\times4=\dfrac{8}{3}$ J.
Q19 — Work and Energy · medium
A force $F$ acting on an object varies with distance $x$ as shown here. The force is in newton and $x$ is in metre. The work done by the force in moving the object from $x=0$ to $x=6$ m is
A. 4.5 J
B. 13.5 J ✓ Correct
C. 9.0 J
D. 18.0 J
Solution: Work done from $x=0$ to $x=6$ m equals the area under the curve $=$ area of square $+$ area of triangle $=3\times3+\dfrac{1}{2}\times3\times3=9+4.5=13.5$ J.
Q20 — Work and Energy · medium
A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is $6\ \text{ms}^{-1}$. The kinetic energy of the other mass is
A. 256 J
B. 486 J ✓ Correct
C. 524 J
D. 324 J
Solution: By conservation of momentum, $m_1u_1=m_2u_2\Rightarrow 18\times6=12\times u_2\Rightarrow u_2=9\ \text{ms}^{-1}$. Kinetic energy of 12 kg mass $=\dfrac{1}{2}m_2u_2^2=\dfrac{1}{2}\times12\times9^2=6\times81=486$ J.
Q21 — Work and Energy · medium
A particle of mass $m_1$ is moving with a velocity $v_1$ and another particle of mass $m_2$ is moving with a velocity $v_2$. Both of them have the same momentum, but their different kinetic energies are $E_1$ and $E_2$ respectively. If $m_1>m_2$, then
A. $E_1<E_2$ ✓ Correct
B. $\dfrac{E_1}{E_2}=\dfrac{m_1}{m_2}$
C. $E_1>E_2$
D. $E_1=E_2$
Solution: $E=\dfrac{1}{2}mv^2=\dfrac{(mv)^2}{2m}=\dfrac{p^2}{2m}$. For the same momentum $p$, $E\propto\dfrac{1}{m}$. Since $m_1>m_2$, $E_1<E_2$.
Q22 — Work and Energy · medium
A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 ft tall building. After a fall of 30 ft each towards earth, their respective kinetic energies will be in the ratio of
A. $\sqrt{2}:1$
B. $1:4$
C. $1:2$ ✓ Correct
D. $1:\sqrt{2}$
Solution: The velocity of a freely falling body does not depend on its mass; it depends only on the height fallen. So $v_1=v_2=v$ after 30 ft. Thus $\dfrac{K_1}{K_2}=\dfrac{\frac{1}{2}m_1v^2}{\frac{1}{2}m_2v^2}=\dfrac{m_1}{m_2}=\dfrac{2}{4}=\dfrac{1}{2}$, i.e. $1:2$.
Q23 — Work and Energy · medium
If kinetic energy of a body is increased by 300%, then percentage change in momentum will be
A. 100% ✓ Correct
B. 150%
C. 265%
D. 73.2%
Solution: $K=\dfrac{p^2}{2m}\Rightarrow p=\sqrt{2mK}$. New KE $K'=K+\dfrac{300}{100}K=4K$, so $p'=\sqrt{2m\times4K}=2\sqrt{2mK}=2p$. Percentage change $=\dfrac{p'-p}{p}\times100=\left(\dfrac{2p}{p}-1\right)\times100=100\%$.
Q24 — Work and Energy · medium
A stone is thrown at an angle of $45^\circ$ to the horizontal with kinetic energy $K$. The kinetic energy at the highest point is
A. $\dfrac{K}{2}$ ✓ Correct
B. $\dfrac{K}{\sqrt{2}}$
C. $K$
D. zero
Solution: At the highest point, $v_x=u\cos\theta$ and $v_y=0$, so $K_H=\dfrac{1}{2}mv_x^2=\dfrac{1}{2}mu^2\cos^2\theta$. Initial KE $K=\dfrac{1}{2}mu^2$, hence $K_H=K\cos^2\theta=K\cos^2 45^\circ=K\times\left(\dfrac{1}{\sqrt{2}}\right)^2=\dfrac{K}{2}$.
Q25 — Work and Energy · medium
A child is swinging a swing. Minimum and maximum heights of swing from the earth's surface are 0.75 m and 2 m respectively. The maximum velocity of this swing is
A. 5 m/s ✓ Correct
B. 10 m/s
C. 15 m/s
D. 20 m/s
Solution: From energy conservation, $\dfrac{1}{2}mv_{\max}^2=mg(H_2-H_1)$. With $H_1=0.75$ m and $H_2=2$ m, $v_{\max}^2=2\times10\times(2-0.75)=2\times10\times1.25=25\Rightarrow v_{\max}=5$ m/s.
Q26 — Work and Energy · medium
Two bodies with kinetic energies in the ratio $4:1$ are moving with equal linear momentum. The ratio of their masses is
A. $1:2$
B. $1:1$
C. $4:1$
D. $1:4$ ✓ Correct
Solution: $\text{KE}=\dfrac{p^2}{2m}$. For equal momentum $p_1=p_2$, $\text{KE}\propto\dfrac{1}{m}$. So $\dfrac{m_1}{m_2}=\dfrac{\text{KE}_2}{\text{KE}_1}=\dfrac{1}{4}$, i.e. $1:4$.
Q27 — Work and Energy · medium
A force acts on a 3.0 g particle in such a way that the position of the particle as a function of time is given by $x=3t-4t^2+t^3$, where $x$ is in metre and $t$ in second. The work done during the first 4 s is
A. 570 mJ
B. 450 mJ
C. 490 mJ
D. 528 mJ ✓ Correct
Solution: $v=\dfrac{dx}{dt}=3-8t+3t^2$. At $t=0$, $v_1=3$ m/s; at $t=4$ s, $v_2=3-32+48=19$ m/s. Work done $=$ gain in KE $=\dfrac{1}{2}m(v_2^2-v_1^2)=\dfrac{1}{2}\times3\times10^{-3}\times(19^2-3^2)=1.5\times10^{-3}\times(19+3)(19-3)=1.5\times10^{-3}\times352=528\times10^{-3}$ J $=528$ mJ.
Q28 — Work and Energy · medium
A rubber ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known. On bouncing it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of
A. $\dfrac{16}{25}$
B. $\dfrac{2}{5}$ ✓ Correct
C. $\dfrac{3}{5}$
D. $\dfrac{9}{25}$
Solution: Using $mgh=\dfrac{1}{2}mv^2\Rightarrow v=\sqrt{2gh}$. Fractional loss in velocity $=\dfrac{\Delta v}{v_1}=\dfrac{\sqrt{2gh_1}-\sqrt{2gh_2}}{\sqrt{2gh_1}}=1-\sqrt{\dfrac{h_2}{h_1}}=1-\sqrt{\dfrac{1.8}{5}}=1-\sqrt{0.36}=1-0.6=0.4=\dfrac{2}{5}$.
Q29 — Work and Energy · medium
If the momentum of a body is increased by 50%, then the percentage increase in its kinetic energy is
A. 50%
B. 100%
C. 125% ✓ Correct
D. 200%
Solution: New momentum $p_2=\dfrac{150}{100}p_1$, so $v_2=\dfrac{15}{10}v_1$. Then $\dfrac{E_2}{E_1}=\left(\dfrac{v_2}{v_1}\right)^2=\left(\dfrac{15}{10}\right)^2=\dfrac{225}{100}$. Percentage increase $=\left(\dfrac{225}{100}-1\right)\times100=125\%$.
Q30 — Work and Energy · medium
The KE acquired by a mass $m$ in travelling a certain distance $d$, starting from rest, under the action of a constant force is directly proportional to
A. $m$
B. $\sqrt{m}$
C. $\dfrac{1}{\sqrt{m}}$
D. Independent of $m$ ✓ Correct
Solution: Starting from rest, $v^2=2ad$ with $a=\dfrac{F}{m}$, so $v^2=2\dfrac{F}{m}d$. Then $KE=\dfrac{1}{2}mv^2=\dfrac{1}{2}m\times2\dfrac{F}{m}d=Fd$, i.e. KE acquired $=F\times d$, which is independent of the mass $m$.