Work Power Energy JEE Advanced previous year questions with solutions

4 solved JEE Advanced questions on Work Power Energy, free to read — no sign-in needed. The full chapter has 20 questions; sign in to attempt the remaining 16 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Advanced Adv 2018 (Paper 2)
   A particle of mass $m$ is initially at rest at the origin. It is subjected to a force and starts moving along the x - axis. Its kinetic energy changes with time as $dK/dt=\gamma t,$ where $\gamma$ is a positive constant of appropriate dimensions. Which of the following statements is (are) true?
   1. A.The force applied on the particle is constant
   2. B.The speed of the particle is proportional to time
   3. C.The distance of the particle from the origin increases linearly with time
   4. D.Since force is constant, it is conservative.
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   Answer: A,B,D

   Given the change in kinetic energy$\dfrac{dk}{dt}=\gamma t$. Kinetic energy is expressed as $k=\dfrac{1}{2}m{v}^{2}$. $∴\dfrac{dk}{dt}=mv\dfrac{dv}{dt}=\gamma t$ Integrating the above relation, $m{\int }_{0}^{v}vdv=\gamma {\int }_{0}^{t}tdt$ $\dfrac{m{v}^{2}}{2}=\dfrac{\gamma {t}^{2}}{2}$ $v=\sqrt{\dfrac{\gamma }{m}}t;υ\propto t$ ... (i) $a=\dfrac{dv}{dt}=\sqrt{\dfrac{\gamma }{m}}=constant$ Since $F=ma$ $∴F=m\sqrt{\dfrac{\gamma }{m}}=\sqrt{\gamma m}=constant$
2. Q2JEE Advanced Adv 2013 (Paper 1)
   A particle of mass $\text{0.2 kg}$ is moving in one dimension under a force that delivers a constant power $\text{0.5 W}$ to the particle. If the initial speed $\left({in ms}^{-1}\right)$ of the particle is zero, the speed $\left({in ms}^{-1}\right)$ after $\text{5 s}$ is
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   Answer: 5

   $\text{Power}=\text{constant}=\text{0.5 w ​}$ $\text{work done w}=\text{pt}$ $=\text{0.5}\times 5=\text{2.5 J}$ $\text{w}=\Delta \text{k.E}=\dfrac{1}{2}\text{m}{\text{v}}^{2}$ $\text{2.5}=\dfrac{1}{2}\times 0\cdot 2{v}^{2}$ $\dfrac{5}{\text{0.2}}={\text{v}}^{2}$ ${\text{v}}^{2}=25$ $\text{v}=5$
3. Q3JEE Advanced Adv 2013 (Paper 1)
   The work done on a particle of mass m by a force $\text{K}\left[\dfrac{\text{x}}{{\left({\text{x}}^{2}+{\text{y}}^{2}\right)}^{3/2}}\hat{\text{i}}+\dfrac{\text{y}}{{\left({\text{x}}^{2}+{\text{y}}^{2}\right)}^{3/2}}\hat{\text{j}}\right]$ (K being a constant of appropriate dimensions, when the particle is taken from the point (a,0) to the point (0,a) along a circular path of radius a about the origin in the x-y plane is
   1. A.$\dfrac{2\text{K}\pi }{\text{a}}$
   2. B.$\dfrac{\text{K}\pi }{\text{a}}$
   3. C.$\dfrac{\text{K}\pi }{2\text{a}}$
   4. D.0
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   Answer: (D)

   $\text{dw}==\vec{\text{F}}\cdot \text{d}\vec{\text{r}}=\vec{\text{F}}\cdot \left(\text{dx}\hat{\text{i}}+\text{dy}\hat{\text{j}}\right)=\text{K}\int \dfrac{\text{xdx}}{{\left({\text{x}}^{2}+{\text{y}}^{2}\right)}^{3/2}}+\dfrac{\text{ydy}}{{\left({\text{x}}^{2}+{\text{y}}^{2}\right)}^{3/2}}$ ​x2 + y2 = a2 $\text{w}=\dfrac{\text{K}}{{\text{a}}^{3}}{\int }_{\text{a}}^{0}xdx+{\int }_{0}^{\text{a}}ydy=\dfrac{\text{K}}{{\text{a}}^{3}}\left(\dfrac{-{\text{a}}^{2}}{2}+\dfrac{{\text{a}}^{2}}{2}\right)=0\text{.}$
4. Q4JEE Advanced Adv 2007 (Paper 1)
   Statement I A block of mass $m$ starts moving on a rough horizontal surface with a velocity $v$. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of $30^{\circ}$ with the horizontal and the same block is made to go up on the surface with the same initial velocity $v$. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. Statement II The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.
   1. A.Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I
   2. B.Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I
   3. C.Statement I is true, Statement II is false
   4. D.Statement I is false, Statement II is true
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   Answer: (C)

   In statement I Decrease in mechanical energy in case I will be $$\Delta U_1=\frac{1}{2} m v^2$$ But decrease in mechanical energy in case II will be $$\begin{aligned} & \Delta U_2=\frac{1}{2} m v^2-m g h \\ & \Delta U_2 < \Delta U_1 \end{aligned}$$ $$\therefore \quad \Delta U_2 < \Delta U_1$$ or Statement $\mathrm{I}$ is correct. In Statement II Coefficient of friction will not change or this Statement is wrong. $\therefore$ Option (c) is correct.