Solved Numericals on Work, Energy & Power

Introduction: Why Numericals Matter in Physics

Physics is a subject built on understanding concepts and applying them to solve real-world problems. While theoretical knowledge provides the foundation, numericals or problem-solving exercises bring these ideas to life. They test understanding, improve analytical skills, and show how abstract formulas translate into real-world quantities.

In mechanics, the concepts of work, energy, and power are crucial for understanding motion, forces, and energy transformation. Solving numericals in these topics helps students:

Visualize energy transfer: Understand how forces do work and change energy.
Apply formulas correctly: Gain fluency in using W=Fdcos⁡θW = Fd \cos \thetaW=Fdcosθ, KE=12mv2KE = \frac{1}{2} mv^2KE=21mv2, PE=mghPE = mghPE=mgh, and P=W/tP = W/tP=W/t.
Prepare for exams: Many competitive and board exams focus heavily on problem-solving.
Bridge theory and application: Learn the practical meaning behind mathematical equations.

By systematically solving problems, students gain confidence and develop a deeper grasp of physics principles.

Concept Recap: Work, Energy, and Power

Before diving into numericals, let’s recall the definitions and formulas.

1. Work

Definition: Work is done when a force is applied to an object and it moves in the direction of the force. W=F⃗⋅d⃗=Fdcos⁡θW = \vec{F} \cdot \vec{d} = F d \cos \thetaW=F⋅d=Fdcosθ

FFF = magnitude of force
ddd = displacement of object
θ\thetaθ = angle between force and displacement

Key Points:

Work is a scalar quantity.
Work can be positive, negative, or zero.
If force is perpendicular to displacement (θ=90∘\theta = 90^\circθ=90∘), W=0W = 0W=0.

2. Energy

Energy is the capacity to do work. Two important forms in mechanics:

a) Kinetic Energy (KE)

KE=12mv2KE = \frac{1}{2} mv^2KE=21mv2

mmm = mass
vvv = velocity

Represents energy due to motion.

b) Potential Energy (PE)

PE=mghPE = mghPE=mgh

Energy due to position in a gravitational field.

3. Power

Power is the rate at which work is done: P=WtP = \frac{W}{t}P=tW

ttt = time
SI unit: Watt (W)

Key Point: Power measures how quickly energy is transferred or converted.

Step-by-Step Solved Numericals

Now, let’s move from basic to advanced examples, showing clear steps and reasoning.

Category 1: Work by Constant Force

Problem 1: A force of 20 N acts on a block, moving it 5 m in the direction of the force. Find the work done.

Solution: W=Fd=20×5=100 JW = Fd = 20 \times 5 = 100\ \text{J}W=Fd=20×5=100 J

Answer: 100 J100\ \text{J}100 J

Explanation: Force and displacement are in the same direction (θ=0\theta=0θ=0), so cos⁡θ=1\cos\theta = 1cosθ=1.

Problem 2: A force of 30 N acts at 60° to the displacement of 4 m. Find work done. W=Fdcos⁡θ=30×4×cos⁡60∘W = Fd \cos \theta = 30 \times 4 \times \cos 60^\circW=Fdcosθ=30×4×cos60∘

cos⁡60∘=0.5\cos 60^\circ = 0.5cos60∘=0.5 W=120×0.5=60 JW = 120 \times 0.5 = 60\ \text{J}W=120×0.5=60 J

Answer: 60 J60\ \text{J}60 J

Category 2: Work Done by Gravity

Problem 3: A 5 kg object falls from a height of 10 m. Find the work done by gravity. W=mgΔh=5×9.8×10=490 JW = mg\Delta h = 5 \times 9.8 \times 10 = 490\ \text{J}W=mgΔh=5×9.8×10=490 J

Answer: 490 J490\ \text{J}490 J (positive, as force and displacement are downward)

Problem 4: A person lifts a 12 kg box to a height of 3 m. Work done by gravity? W=−mgh=−12×9.8×3=−352.8 JW = -mgh = -12 \times 9.8 \times 3 = -352.8\ \text{J}W=−mgh=−12×9.8×3=−352.8 J

Negative because the displacement is opposite to the direction of gravitational force.

Category 3: Work on an Inclined Plane

Problem 5: A block of 10 kg slides down a frictionless incline of height 2 m. Find work done by gravity. W=mgΔh=10×9.8×2=196 JW = mg\Delta h = 10 \times 9.8 \times 2 = 196\ \text{J}W=mgΔh=10×9.8×2=196 J

Note: The work depends only on vertical height, not on the slope length.

Category 4: Work by a Variable Force

Problem 6: A particle experiences F(x)=5x2F(x) = 5x^2F(x)=5×2 along x-axis. Calculate work from x=0x=0x=0 to x=3x=3x=3 m. W=∫035x2dx=5[x33]03=5×9=45 JW = \int_0^3 5x^2 dx = 5 \left[ \frac{x^3}{3} \right]_0^3 = 5 \times 9 = 45\ \text{J}W=∫035x2dx=5[3×3]03=5×9=45 J

Answer: 45 J45\ \text{J}45 J

Category 5: Elastic Potential Energy

Problem 7: Spring constant k=200 N/mk = 200\ \text{N/m}k=200 N/m, compressed by 0.1 m. Find stored energy. U=12kx2=0.5×200×0.01=1 JU = \frac{1}{2} k x^2 = 0.5 \times 200 \times 0.01 = 1\ \text{J}U=21kx2=0.5×200×0.01=1 J

Category 6: Kinetic Energy & Work–Energy Theorem

Problem 8: A 2 kg object accelerates from 5 m/s to 15 m/s. Find the work done. W=ΔKE=12m(vf2−vi2)=0.5×2(225−25)=200 JW = \Delta KE = \frac{1}{2} m (v_f^2 – v_i^2) = 0.5 \times 2 (225 – 25) = 200\ \text{J}W=ΔKE=21m(vf2−vi2)=0.5×2(225−25)=200 J

Answer: 200 J200\ \text{J}200 J

Category 7: Power

Problem 9: A car of mass 1000 kg accelerates from rest to 20 m/s in 10 s. Find average power. KE=12mv2=0.5×1000×400=200,000 JKE = \frac{1}{2} mv^2 = 0.5 \times 1000 \times 400 = 200,000\ \text{J}KE=21mv2=0.5×1000×400=200,000 J Pavg=KEt=200,00010=20,000 W=20 kWP_{avg} = \frac{KE}{t} = \frac{200,000}{10} = 20,000\ \text{W} = 20\ \text{kW}Pavg=tKE=10200,000=20,000 W=20 kW

Problem 10: Lifting a 50 kg person 10 m in 20 s. Power? W=mgh=50×9.8×10=4900 JW = mgh = 50 \times 9.8 \times 10 = 4900\ \text{J}W=mgh=50×9.8×10=4900 J P=W/t=4900/20=245 WP = W/t = 4900 / 20 = 245\ \text{W}P=W/t=4900/20=245 W

Category 8: Work Against Friction

Problem 11: A 15 kg box is pulled 5 m on a rough surface with μ=0.2\mu=0.2μ=0.2, g=9.8g=9.8g=9.8. Find work done against friction. Ff=μmg=0.2×15×9.8=29.4 NF_f = \mu mg = 0.2 \times 15 \times 9.8 = 29.4\ \text{N}Ff=μmg=0.2×15×9.8=29.4 N Wf=Ffd=29.4×5=147 JW_f = F_f d = 29.4 \times 5 = 147\ \text{J}Wf=Ffd=29.4×5=147 J

Answer: 147 J147\ \text{J}147 J

Category 9: Combined Work & Power

Problem 12: A 80 kg person climbs 12 m stairs in 15 s. Power delivered? W=mgh=80×9.8×12=9,408 JW = mgh = 80 \times 9.8 \times 12 = 9,408\ \text{J}W=mgh=80×9.8×12=9,408 J P=W/t=9,408/15≈627.2 WP = W/t = 9,408 / 15 \approx 627.2\ \text{W}P=W/t=9,408/15≈627.2 W

Category 10: Rotational Work

Problem 13: A torque of 10 N·m rotates a shaft through 20 revolutions. Work done? θ=20×2π=40π rad\theta = 20 \times 2\pi = 40\pi\ \text{rad}θ=20×2π=40π rad W=τθ=10×40π=400π≈1256.64 JW = \tau \theta = 10 \times 40\pi = 400\pi \approx 1256.64\ \text{J}W=τθ=10×40π=400π≈1256.64 J

Discussion of Results

Direction Matters: Positive or negative work depends on the relative direction of force and displacement.
Path Independence for Conservative Forces: Gravity and spring forces only depend on vertical height or compression, not actual path.
Energy Conversion: Work done by forces often converts between kinetic and potential energy.
Power Interpretation: Gives rate of energy transfer; high power means faster work or energy conversion.
Friction Reduces Efficiency: Work against friction does not increase kinetic energy but is dissipated as heat.

Practice Tips

Identify forces carefully: Separate gravity, friction, applied forces.
Check angles: Use cos⁡θ\cos\thetacosθ correctly in work calculations.
Use work–energy theorem: Sometimes faster than integrating forces.
Sign conventions: Upwards against gravity = negative work; downwards = positive.
Units consistency: Newtons, meters, seconds.