📘 Work Power Energy — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Work Power Energy for JEE Main with previous year questions on the work-energy theorem, power, and spring systems. Each solution is explained in a simple, step-by-step manner to boost your score.
Work Power Energy — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (3 April Shift 1)
A particle is released from height \(S\) above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively :
(1) \(\frac{S}{2}, \sqrt{\frac{3gS}{2}}\) (2) \(\frac{S}{2}, \frac{3gS}{2}\)
(3) \(\frac{S}{4}, \frac{3gS}{2}\) (4) \(\frac{S}{4}, \sqrt{\frac{3gS}{2}}\)
(1) \(\frac{S}{2}, \sqrt{\frac{3gS}{2}}\) (2) \(\frac{S}{2}, \frac{3gS}{2}\)
(3) \(\frac{S}{4}, \frac{3gS}{2}\) (4) \(\frac{S}{4}, \sqrt{\frac{3gS}{2}}\)
✅ Answer & Explanation:
Correct Option: (4)
Total Energy at height \(S = mgS\).
Let height be \(h\). Given \(KE = 3PE \implies KE = 3mgh\).
By Conservation of Energy: \(KE + PE = mgS \implies 3mgh + mgh = mgS \implies 4mgh = mgS \implies h = S/4\).
Speed: \(KE = \frac{1}{2}mv^2 = 3mg(S/4) \implies v^2 = \frac{3gS}{2} \implies v = \sqrt{\frac{3gS}{2}}\).
Total Energy at height \(S = mgS\).
Let height be \(h\). Given \(KE = 3PE \implies KE = 3mgh\).
By Conservation of Energy: \(KE + PE = mgS \implies 3mgh + mgh = mgS \implies 4mgh = mgS \implies h = S/4\).
Speed: \(KE = \frac{1}{2}mv^2 = 3mg(S/4) \implies v^2 = \frac{3gS}{2} \implies v = \sqrt{\frac{3gS}{2}}\).
Q2. JEE Main 2025 (23 Jan Shift 2)
A ball having kinetic energy KE, is projected at an angle of \(60^\circ\) from the horizontal. What will be the kinetic energy of ball at the highest point of its flight ?
(1) \(\frac{KE}{8}\) (2) \(\frac{KE}{2}\)
(3) \(\frac{KE}{16}\) (4) \(\frac{KE}{4}\)
(1) \(\frac{KE}{8}\) (2) \(\frac{KE}{2}\)
(3) \(\frac{KE}{16}\) (4) \(\frac{KE}{4}\)
✅ Answer & Explanation:
Correct Option: (4)
At projection: \(KE = \frac{1}{2}mu^2\).
At highest point, only horizontal velocity remains: \(v = u \cos 60^\circ = u/2\).
\(KE_{highest} = \frac{1}{2}m(u/2)^2 = \frac{1}{4}(\frac{1}{2}mu^2) = \frac{KE}{4}\).
At projection: \(KE = \frac{1}{2}mu^2\).
At highest point, only horizontal velocity remains: \(v = u \cos 60^\circ = u/2\).
\(KE_{highest} = \frac{1}{2}m(u/2)^2 = \frac{1}{4}(\frac{1}{2}mu^2) = \frac{KE}{4}\).
Q3. JEE Main 2025 (22 Jan Shift 2)
A force \(\vec{F} = 2\hat{i} + b\hat{j} + \hat{k}\) is applied on a particle and it undergoes a displacement \(\hat{i} - 2\hat{j} - \hat{k}\). What will be the value of \(b\), if work done on the particle is zero?
(1) 0 (2) \(\frac{1}{2}\)
(3) 2 (4) \(\frac{1}{3}\)
(1) 0 (2) \(\frac{1}{2}\)
(3) 2 (4) \(\frac{1}{3}\)
✅ Answer & Explanation:
Correct Option: (2)
Work done \(W = \vec{F} \cdot \vec{d} = 0\).
\((2\hat{i} + b\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} - \hat{k}) = 0\)
\(2(1) + b(-2) + 1(-1) = 0 \implies 2 - 2b - 1 = 0 \implies 1 = 2b \implies b = 1/2\).
Work done \(W = \vec{F} \cdot \vec{d} = 0\).
\((2\hat{i} + b\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} - \hat{k}) = 0\)
\(2(1) + b(-2) + 1(-1) = 0 \implies 2 - 2b - 1 = 0 \implies 1 = 2b \implies b = 1/2\).
Q4. JEE Main 2024 (04 Apr Shift 2)
A body of \(m \, \text{kg}\) slides from rest along the curve of vertical circle from point A to B in friction less path. The velocity of the body at B is : (given, \(R = 14 \, \text{m}, g = 10 \, \text{m/s}^2\) and \(\sqrt{2} = 1.4\))
(1) 16.7 m/s (2) 19.8 m/s
(3) 10.6 m/s (4) 21.9 m/s
(3) 10.6 m/s (4) 21.9 m/s
✅ Answer & Explanation:
Correct Option: (4)
Height difference \(h = R + R \sin 45^\circ = 14(1 + 0.7) = 23.8 \, \text{m}\).
Velocity \(v = \sqrt{2gh} = \sqrt{2 \times 10 \times 23.8} = \sqrt{476} \approx 21.9 \, \text{m/s}\).
Height difference \(h = R + R \sin 45^\circ = 14(1 + 0.7) = 23.8 \, \text{m}\).
Velocity \(v = \sqrt{2gh} = \sqrt{2 \times 10 \times 23.8} = \sqrt{476} \approx 21.9 \, \text{m/s}\).
Q5. JEE Main 2024 (04 Apr Shift 1)
If a rubber ball falls from a height \(h\) and rebounds upto the height of \(h/2\). The percentage loss of total energy of the initial system as well as velocity ball before it strikes the ground, respectively, are :
(1) 50%, \(\sqrt{2gh}\) (2) 50%, \(\sqrt{gh}\)
(3) 40%, \(\sqrt{2gh}\) (4) 50%, \(\sqrt{\frac{gh}{2}}\)
(1) 50%, \(\sqrt{2gh}\) (2) 50%, \(\sqrt{gh}\)
(3) 40%, \(\sqrt{2gh}\) (4) 50%, \(\sqrt{\frac{gh}{2}}\)
✅ Answer & Explanation:
Correct Option: (1)
- Initial Energy = \(mgh\). Final Energy = \(mg(h/2)\). Loss = \(mgh/2 = 50\%\).
- Velocity just before strike: \(v = \sqrt{2gh}\).
- Initial Energy = \(mgh\). Final Energy = \(mg(h/2)\). Loss = \(mgh/2 = 50\%\).
- Velocity just before strike: \(v = \sqrt{2gh}\).
Q6. JEE Main 2022 (25 Jul Shift 1)
A body of mass 0.5 kg travels on straight line path with velocity \(v = (3x^2 + 4) \, \text{m/s}\). The net work done by the force during its displacement from \(x = 0\) to \(x = 2 \, \text{m}\) is :
(1) 64 J (2) 60 J
(3) 120 J (4) 128 J
(1) 64 J (2) 60 J
(3) 120 J (4) 128 J
✅ Answer & Explanation:
Correct Option: (2)
Work done = Change in KE.
- At \(x = 0\), \(v_1 = 4 \, \text{m/s}\).
- At \(x = 2\), \(v_2 = 3(2^2) + 4 = 16 \, \text{m/s}\).
\(W = \frac{1}{2}m(v_2^2 - v_1^2) = \frac{1}{2}(0.5)(256 - 16) = 0.25 \times 240 = 60 \, \text{J}\).
Work done = Change in KE.
- At \(x = 0\), \(v_1 = 4 \, \text{m/s}\).
- At \(x = 2\), \(v_2 = 3(2^2) + 4 = 16 \, \text{m/s}\).
\(W = \frac{1}{2}m(v_2^2 - v_1^2) = \frac{1}{2}(0.5)(256 - 16) = 0.25 \times 240 = 60 \, \text{J}\).
Q7. JEE Main 2021 (18 Mar Shift 1)
As shown in the figure, a particle of mass \(m\) is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is \(v\). (Take \(g = 10 \, \text{m/s}^2\)). The value of \(v\) to the nearest integer is _____ m/s.
✅ Answer & Explanation:
Correct Answer: 10
Using conservation of energy between point A (height 10m) and point B (height 5m):
\(mg(10) = mg(5) + \frac{1}{2}mv^2 \implies 5mg = \frac{1}{2}mv^2 \implies v = \sqrt{10g} = \sqrt{100} = 10 \, \text{m/s}\).
Using conservation of energy between point A (height 10m) and point B (height 5m):
\(mg(10) = mg(5) + \frac{1}{2}mv^2 \implies 5mg = \frac{1}{2}mv^2 \implies v = \sqrt{10g} = \sqrt{100} = 10 \, \text{m/s}\).
Q8. JEE Main 2025 (29 Jan Shift 2)
A sand dropper drops sand of mass \(m(t)\) on a conveyer belt at a rate proportional to the square root of speed \((v)\) of the belt, i.e. \(\frac{dm}{dt} \propto \sqrt{v}\). If \(P\) is the power delivered to run the belt at constant speed then which of the following relationship is true?
(1) \(P \propto \sqrt{v}\) (2) \(P \propto v\)
(3) \(P^2 \propto v^5\) (4) \(P^2 \propto v^3\)
(1) \(P \propto \sqrt{v}\) (2) \(P \propto v\)
(3) \(P^2 \propto v^5\) (4) \(P^2 \propto v^3\)
✅ Answer & Explanation:
Correct Option: (3)
Force needed to keep belt at constant speed: \(F = v \frac{dm}{dt}\).
Power \(P = Fv = v^2 \frac{dm}{dt}\).
Given \(\frac{dm}{dt} = k\sqrt{v}\).
\(P = v^2 (k\sqrt{v}) = kv^{2.5} \implies P^2 \propto v^5\).
Force needed to keep belt at constant speed: \(F = v \frac{dm}{dt}\).
Power \(P = Fv = v^2 \frac{dm}{dt}\).
Given \(\frac{dm}{dt} = k\sqrt{v}\).
\(P = v^2 (k\sqrt{v}) = kv^{2.5} \implies P^2 \propto v^5\).
Q9. JEE Main 2025 (28 Jan Shift 1)
Assertion A: In a central force field, the work done is independent of the path chosen.
Reason R: Every force encountered in mechanics does not have an associated potential energy.
Choose the most appropriate answer:
(1) A is false but R is true
(2) Both A and R are true but R is NOT the correct explanation of A
(3) A is true but R is false
(4) Both A and R are true and R is the correct explanation of A
Reason R: Every force encountered in mechanics does not have an associated potential energy.
Choose the most appropriate answer:
(1) A is false but R is true
(2) Both A and R are true but R is NOT the correct explanation of A
(3) A is true but R is false
(4) Both A and R are true and R is the correct explanation of A
✅ Answer & Explanation:
Correct Option: (2)
- Assertion: True. Central forces (like gravity) are conservative, so work done is path-independent.
- Reason: True. Non-conservative forces (like friction) do not have an associated potential energy. However, this fact does not explain why central forces specifically are path-independent.
- Assertion: True. Central forces (like gravity) are conservative, so work done is path-independent.
- Reason: True. Non-conservative forces (like friction) do not have an associated potential energy. However, this fact does not explain why central forces specifically are path-independent.
Q10. JEE Main 2024 (27 Jan Shift 2)
A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle (\(\theta\)) of thread deflection in the extreme position will be :
(1) \(\tan^{-1}(\sqrt{2})\) (2) \(2 \tan^{-1}(\frac{1}{2})\)
(3) \(\tan^{-1}(\frac{1}{2})\) (4) \(2 \tan^{-1}(\frac{1}{\sqrt{5}})\)
(1) \(\tan^{-1}(\sqrt{2})\) (2) \(2 \tan^{-1}(\frac{1}{2})\)
(3) \(\tan^{-1}(\frac{1}{2})\) (4) \(2 \tan^{-1}(\frac{1}{\sqrt{5}})\)
✅ Answer & Explanation:
Correct Option: (4)
Solving the equality of extreme tangential acceleration (\(g \sin\theta\)) and lowest centripetal acceleration (\(v^2/R\)) yields \(\theta = 2 \tan^{-1}(1/\sqrt{5})\).
Solving the equality of extreme tangential acceleration (\(g \sin\theta\)) and lowest centripetal acceleration (\(v^2/R\)) yields \(\theta = 2 \tan^{-1}(1/\sqrt{5})\).
Q11. JEE Main 2024 (05 Apr Shift 2)
A body is moving unidirectionally under the influence of a constant power source. Its displacement in time \(t\) is proportional to :
(1) \(t\) (2) \(t^{3/2}\)
(3) \(t^2\) (4) \(t^{2/3}\)
(1) \(t\) (2) \(t^{3/2}\)
(3) \(t^2\) (4) \(t^{2/3}\)
✅ Answer & Explanation:
Correct Option: (2)
Power \(P = Fv = m \frac{dv}{dt} v = \text{constant}\).
\(\int v dv = \int \frac{P}{m} dt \implies v^2 \propto t \implies v \propto \sqrt{t}\).
Displacement \(s = \int v dt \propto \int t^{1/2} dt \implies s \propto t^{3/2}\).
Power \(P = Fv = m \frac{dv}{dt} v = \text{constant}\).
\(\int v dv = \int \frac{P}{m} dt \implies v^2 \propto t \implies v \propto \sqrt{t}\).
Displacement \(s = \int v dt \propto \int t^{1/2} dt \implies s \propto t^{3/2}\).
Q12. JEE Main 2022 (24 Jun Shift 2)
Potential energy as a function of \(r\) is given by \(U = \frac{A}{r^{10}} - \frac{B}{r^5}\), where \(r\) is the interatomic distance, A and B are positive constants. The equilibrium distance between the two atoms will be :
(1) \((\frac{A}{B})^{1/5}\) (2) \((\frac{B}{A})^{1/5}\)
(3) \((\frac{2A}{B})^{1/5}\) (4) \((\frac{B}{2A})^{1/5}\)
(1) \((\frac{A}{B})^{1/5}\) (2) \((\frac{B}{A})^{1/5}\)
(3) \((\frac{2A}{B})^{1/5}\) (4) \((\frac{B}{2A})^{1/5}\)
✅ Answer & Explanation:
Correct Option: (3)
At equilibrium, Force \(F = -dU/dr = 0\).
\(- [ \frac{-10A}{r^{11}} + \frac{5B}{r^6} ] = 0 \implies \frac{10A}{r^{11}} = \frac{5B}{r^6} \implies r^5 = \frac{2A}{B}\).
\(r = (2A/B)^{1/5}\).
At equilibrium, Force \(F = -dU/dr = 0\).
\(- [ \frac{-10A}{r^{11}} + \frac{5B}{r^6} ] = 0 \implies \frac{10A}{r^{11}} = \frac{5B}{r^6} \implies r^5 = \frac{2A}{B}\).
\(r = (2A/B)^{1/5}\).
Q13. JEE Main 2021 (25 Feb Shift 1)
The potential energy \(U\) of a diatomic molecule is a function dependent on \(r\) (interatomic distance) as \(U = \frac{A}{r^{12}} - \frac{B}{r^6}\) where \(A\) and \(B\) are positive constants. The equilibrium distance between two atoms will be \((\frac{2A}{B})^{1/6}\). True (1) or False (2)?
✅ Answer & Explanation:
Correct Answer: 1 (True)
Setting \(dU/dr = 0 \implies \frac{-12A}{r^{13}} + \frac{6B}{r^7} = 0 \implies \frac{12A}{r^{13}} = \frac{6B}{r^7} \implies r^6 = \frac{2A}{B}\).
Setting \(dU/dr = 0 \implies \frac{-12A}{r^{13}} + \frac{6B}{r^7} = 0 \implies \frac{12A}{r^{13}} = \frac{6B}{r^7} \implies r^6 = \frac{2A}{B}\).
Q14. JEE Main 2025 (29 Jan Shift 1)
A body of mass '\(m\)' connected to a massless and unstretchable string goes in verticle circle of radius '\(R\)' under gravity \(g\). The other end of the string is fixed at the center of circle. If velocity at top of circular path is \(n\sqrt{gR}\), where, \(n \ge 1\), then ratio of kinetic energy of the body at bottom to that at top of the circle is :
(1) \(\frac{n^2}{n^2+4}\) (2) \(\frac{n^2+4}{n^2}\)
(3) \(\frac{n+4}{n}\) (4) \(\frac{n}{n+4}\)
(1) \(\frac{n^2}{n^2+4}\) (2) \(\frac{n^2+4}{n^2}\)
(3) \(\frac{n+4}{n}\) (4) \(\frac{n}{n+4}\)
✅ Answer & Explanation:
Correct Option: (2)
\(KE_{top} = \frac{1}{2}m(n\sqrt{gR})^2 = \frac{1}{2}mn^2gR\).
By conservation of energy: \(KE_{bottom} = KE_{top} + mg(2R) = \frac{1}{2}mn^2gR + 2mgR = \frac{1}{2}mgR(n^2 + 4)\).
Ratio \(\frac{KE_{bottom}}{KE_{top}} = \frac{\frac{1}{2}mgR(n^2+4)}{\frac{1}{2}mn^2gR} = \frac{n^2+4}{n^2}\).
\(KE_{top} = \frac{1}{2}m(n\sqrt{gR})^2 = \frac{1}{2}mn^2gR\).
By conservation of energy: \(KE_{bottom} = KE_{top} + mg(2R) = \frac{1}{2}mn^2gR + 2mgR = \frac{1}{2}mgR(n^2 + 4)\).
Ratio \(\frac{KE_{bottom}}{KE_{top}} = \frac{\frac{1}{2}mgR(n^2+4)}{\frac{1}{2}mn^2gR} = \frac{n^2+4}{n^2}\).
Q15. JEE Main 2024 (08 Apr Shift 2)
A circular table is rotating with an angular velocity of \(\omega \, \text{rad/s}\) about its axis. There is a smooth groove along a radial direction on the table. A steel ball is gently placed at a distance of 1 m on the groove. If the radius of the table is 3 m, the radial velocity of the ball w.r.t. the table at the time ball leaves the table is \(x\sqrt{2}\omega \, \text{m/s}\), where the value of \(x\) is :
✅ Answer & Explanation:
Correct Answer: 2
Using Work-Energy theorem in rotating frame (centrifugal force):
\(\int_1^3 m \omega^2 r dr = \frac{1}{2}m v_r^2\)
\(\omega^2 [\frac{r^2}{2}]_1^3 = \frac{1}{2}v_r^2 \implies \omega^2(9 - 1) = v_r^2 \implies v_r = \sqrt{8}\omega = 2\sqrt{2}\omega\).
Comparing with \(x\sqrt{2}\omega\), we get \(x = 2\).
Using Work-Energy theorem in rotating frame (centrifugal force):
\(\int_1^3 m \omega^2 r dr = \frac{1}{2}m v_r^2\)
\(\omega^2 [\frac{r^2}{2}]_1^3 = \frac{1}{2}v_r^2 \implies \omega^2(9 - 1) = v_r^2 \implies v_r = \sqrt{8}\omega = 2\sqrt{2}\omega\).
Comparing with \(x\sqrt{2}\omega\), we get \(x = 2\).
Q16. JEE Main 2022 (26 Jun Shift 2)
Arrange the four graphs in descending order of total work done; where \(W_1, W_2, W_3\) and \(W_4\) are the work done corresponding to figure a, b, c and d respectively.
(1) \(W_3 > W_2 > W_1 > W_4\) (2) \(W_3 > W_2 > W_4 > W_1\)
(3) \(W_2 > W_3 > W_4 > W_1\) (4) \(W_2 > W_3 > W_1 > W_4\)
(3) \(W_2 > W_3 > W_4 > W_1\) (4) \(W_2 > W_3 > W_1 > W_4\)
✅ Answer & Explanation:
Correct Option: (1)
Work done is the area under the Force-displacement graph. Comparing the enclosed areas for each case yields the sequence in Option 1.
Work done is the area under the Force-displacement graph. Comparing the enclosed areas for each case yields the sequence in Option 1.
Q17. JEE Main 2022 (24 Jun Shift 1)
A ball of mass 2 kg is dropped from a height 2 m on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance 0.1 m. The spring constant is _____ \(N/m\) (Use \(g = 10 \, \text{m/s}^2\))
✅ Answer & Explanation:
Correct Answer: 8400
Loss in GPE = Gain in SPE
\(mg(h+x) = \frac{1}{2}kx^2\)
\(2 \times 10 \times (2 + 0.1) = \frac{1}{2}k(0.1)^2 \implies 42 = 0.005k \implies k = 8400 \, \text{N/m}\). *Note: Please verify with official JEE Main answer key for this specific year.*
Loss in GPE = Gain in SPE
\(mg(h+x) = \frac{1}{2}kx^2\)
\(2 \times 10 \times (2 + 0.1) = \frac{1}{2}k(0.1)^2 \implies 42 = 0.005k \implies k = 8400 \, \text{N/m}\). *Note: Please verify with official JEE Main answer key for this specific year.*
📘 Exam Preparation Tip for Work, Power & Energy
- Work-Energy Theorem: \(W_{net} = \Delta KE\). This is the most powerful tool. Use it when you need to find work done or change in speed without worrying about time or path details.
- Conservation of Mechanical Energy: \(KE + PE = \text{constant}\) (when only conservative forces act). This is applicable for gravity, spring forces, and other conservative fields.
- Power: \(P = \vec{F} \cdot \vec{v} = \frac{dW}{dt}\). For variable mass systems (like sand on a conveyor belt), remember \(F = v \frac{dm}{dt}\) and \(P = v^2 \frac{dm}{dt}\).
- Potential Energy Curves: For problems with \(U(x)\), use \(F = -\frac{dU}{dx}\). Equilibrium points occur where \(dU/dx = 0\). Stability depends on the second derivative.
- Vertical Circular Motion: At the top, \(T + mg = mv^2/R\). At the bottom, \(T - mg = mv^2/R\). Energy conservation connects speeds at different points.
- Spring Systems: For spring problems, use \(\frac{1}{2}kx^2\) for potential energy. In vertical springs, include gravitational potential energy changes as well.
- Work from Graphs: Work done is the area under the \(F\)-\(x\) curve. For variable forces, integrate or find the area geometrically.