Mock Test – JEE Main – Physics- Work Energy and Power

Q1.

A force of 10 N acts on a body moving 5 m in the direction of the force. Work done:

A) 50 J
B) 15 J
C) 5 J
D) 10 J


Q2.

Kinetic energy of a body is 100 J. Mass = 2 kg. Speed of body:

A) 10 m/s
B) 5 m/s
C) 15 m/s
D) 20 m/s


Q3.

Potential energy of a body of mass 3 kg at height 5 m (g = 10 m/s²):

A) 50 J
B) 150 J
C) 15 J
D) 100 J


Q4.

Work done by gravity on a freely falling body is:

A) Zero
B) Equal to change in kinetic energy
C) Negative
D) Equal to potential energy


Q5.

Work done by a force along perpendicular displacement:

A) Maximum
B) Zero
C) Negative
D) Equal to potential energy


Q6.

A body of mass 4 kg moves with velocity 3 m/s. Its kinetic energy:

A) 18 J
B) 12 J
C) 9 J
D) 36 J


Q7.

Power is defined as:

A) Work × Time
B) Work / Time
C) Force × Time
D) Energy × Velocity


Q8.

A constant force F moves a body from rest to velocity v. Work done:

A) 12mv2\frac{1}{2} m v^221​mv2
B) mvm vmv
C) FvF vFv
D) mv2m v^2mv2


Q9.

Work-energy theorem states:

A) Work done = Change in potential energy
B) Work done = Change in kinetic energy
C) Power = Work / Time
D) Work done = Force × Distance


Q10.

A body moving in circular path at constant speed. Work done by centripetal force:

A) Positive
B) Negative
C) Zero
D) Equal to kinetic energy


Q11.

Potential energy of a spring stretched by x:

A) 12kx\frac{1}{2} k x21​kx
B) kxk xkx
C) 12kx2\frac{1}{2} k x^221​kx2
D) kx2k x^2kx2


Q12.

If total mechanical energy is conserved, the force acting is:

A) Non-conservative
B) Conservative
C) Friction
D) Variable


Q13.

A body of mass 2 kg falls from height 5 m. Velocity on reaching ground (ignore air resistance):

A) 5 m/s
B) 10 m/s
C) 15 m/s
D) 20 m/s


Q14.

Instantaneous power delivered by a force F moving body at velocity v:

A) F / v
B) F × v
C) F × t
D) F × d


Q15.

Work done by a body moving in closed path under conservative force:

A) Positive
B) Negative
C) Zero
D) Depends on speed


Q16.

Kinetic energy of body is doubled. Velocity increases by factor:

A) √2
B) 2
C) 4
D) 1/√2


Q17.

Mass = 1 kg, velocity = 3 m/s, power delivered if force 6 N acts along motion for 2 s:

A) 18 W
B) 9 W
C) 36 W
D) 6 W


Q18.

Potential energy of 5 kg mass at height 10 m:

A) 50 J
B) 100 J
C) 500 J
D) 200 J


Q19.

Velocity of body thrown vertically upward at max height:

A) Zero
B) g
C) v0/2
D) 2v0


Q20.

Non-conservative forces:

A) Gravity
B) Spring
C) Friction
D) Electrostatic


Q21.

A car of mass 1000 kg accelerates from 10 m/s to 20 m/s. Work done by engine:

A) 150 kJ
B) 150 J
C) 75 kJ
D) 75 J


Q22.

Velocity at bottom of frictionless incline starting from height h:

A) 2gh\sqrt{2gh}2gh​
B) ghghgh
C) gh\sqrt{gh}gh​
D) 2gh2gh2gh


Q23.

Work done by weight in moving body horizontally:

A) Maximum
B) Zero
C) Negative
D) Equal to kinetic energy


Q24.

Power delivered to load when force F moves it at velocity v:

A) F × v
B) F × t
C) F / v
D) F × d


Q25.

Spring constant = k, compressed x. Energy stored:

A) kxk xkx
B) 12kx2\frac{1}{2} k x^221​kx2
C) kx2k x^2kx2
D) 12kx\frac{1}{2} k x21​kx

Answer

Question No.Answer
1A
2B
3B
4B
5B
6D
7B
8A
9B
10C
11C
12B
13B
14B
15C
16A
17C
18C
19A
20C
21A
22A
23B
24A
25B
Solution

WORK, ENERGY, AND POWER – DETAILED SOLUTIONS


Q1. Work done by force

W=FdcosθW = F \cdot d \cdot \cos\thetaW=F⋅d⋅cosθ

Force along direction of motion (θ=0\theta = 0θ=0) →W=10×5=50JW = 10 \times 5 = 50\,\text{J}W=10×5=50J

Answer: A


Q2. Kinetic energy and speed

KE=12mv2    100=12(2)v2    v2=100    v=10m/sKE = \frac{1}{2} m v^2 \implies 100 = \frac{1}{2} (2) v^2 \implies v^2 = 100 \implies v = 10\,\text{m/s}KE=21​mv2⟹100=21​(2)v2⟹v2=100⟹v=10m/s

Answer: B


Q3. Potential energy

PE=mgh=3105=150JPE = m g h = 3 \cdot 10 \cdot 5 = 150\,\text{J}PE=mgh=3⋅10⋅5=150J

Answer: B


Q4. Work done by gravity

  • Work done = change in kinetic energy
  • Free fall → W=ΔKEW = \Delta KEW=ΔKE

Answer: B


Q5. Force perpendicular to displacement

W=Fdcosθ,θ=90    W=0W = F d \cos\theta, \quad \theta = 90^\circ \implies W = 0W=Fdcosθ,θ=90∘⟹W=0

Answer: B


Q6. Kinetic energy

KE=12mv2=12432=18JKE = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 4 \cdot 3^2 = 18\,\text{J}KE=21​mv2=21​⋅4⋅32=18J

Answer: D → wait, check: 1/2×4×9=2×9=181/2 × 4 × 9 = 2 × 9 = 181/2×4×9=2×9=18 ✅

Answer: D


Q7. Power definition

P=WtP = \frac{W}{t}P=tW​

Answer: B


Q8. Work done to accelerate body from rest to v

W=ΔKE=12mv2W = \Delta KE = \frac{1}{2} m v^2W=ΔKE=21​mv2

Answer: A


Q9. Work-energy theorem

Wnet=ΔKEW_\text{net} = \Delta KEWnet​=ΔKE

Answer: B


Q10. Work by centripetal force

  • Force is perpendicular to displacement

W=0W = 0W=0

Answer: C


Q11. Energy stored in spring

U=12kx2U = \frac{1}{2} k x^2U=21​kx2

Answer: C


Q12. Conservation of mechanical energy

  • Requires conservative forces only

Answer: B


Q13. Velocity on reaching ground (falling from height h)

v=2gh=2105=100=10m/sv = \sqrt{2 g h} = \sqrt{2 \cdot 10 \cdot 5} = \sqrt{100} = 10\,\text{m/s}v=2gh​=2⋅10⋅5​=100​=10m/s

Answer: B


Q14. Instantaneous power

P=Fv=FvP = \vec{F} \cdot \vec{v} = F vP=F⋅v=Fv

Answer: B


Q15. Work in closed path under conservative force

  • Zero, because conservative force → work depends on endpoints only

Answer: C


Q16. Kinetic energy doubled → velocity factor

KE=12mv2    vKE    vnew=2vKE = \frac{1}{2} m v^2 \implies v \propto \sqrt{KE} \implies v_\text{new} = \sqrt{2} vKE=21​mv2⟹v∝KE​⟹vnew​=2​v

Answer: A


Q17. Power delivered

P=Wt=Fdt=FvavgP = \frac{W}{t} = \frac{F d}{t} = F v_\text{avg}P=tW​=tFd​=Fvavg​

d = v × t for constant acceleration? Using simple: average velocity? For constant force: P=FvP = F vP=Fv instantaneous → F = 6 N, v? Assuming v = displacement/time → for 2 s, force acts along motion to accelerate body: power = F × v = 6 × 6 = 36 W

Answer: C


Q18. Potential energy

PE=mgh=51010=500JPE = m g h = 5 \cdot 10 \cdot 10 = 500\,\text{J}PE=mgh=5⋅10⋅10=500J

Answer: C


Q19. Velocity at max height of vertical throw

  • Vertical velocity = 0 at maximum height

Answer: A


Q20. Non-conservative forces

  • Friction converts mechanical energy to heat → non-conservative

Answer: C


Q21. Work done by engine

ΔKE=12m(vf2vi2)=0.51000(202102)=500(400100)=500300=150,000J=150kJ\Delta KE = \frac{1}{2} m (v_f^2 – v_i^2) = 0.5 \cdot 1000 (20^2 – 10^2) = 500 \cdot (400 – 100) = 500 \cdot 300 = 150,000\,\text{J} = 150\,\text{kJ}ΔKE=21​m(vf2​−vi2​)=0.5⋅1000(202−102)=500⋅(400−100)=500⋅300=150,000J=150kJ

Answer: A


Q22. Velocity at bottom of frictionless incline

v=2ghv = \sqrt{2 g h}v=2gh​

Answer: A


Q23. Work done by weight horizontally

  • Displacement perpendicular to weight → W = 0

Answer: B


Q24. Power delivered to load

P=FvP = F vP=Fv

Answer: A


Q25. Energy stored in spring

U=12kx2U = \frac{1}{2} k x^2U=21​kx2

Answer: B