Work and Energy – Numerical Problems with Stepwise Solutions
CBSE Class 9
Physics — Chapter 11: Work and Energy
Content Bank — Important Formulas (Chapter 11)
Work:
W = \vec{F} · \vec{s} = F s cosθ — only component of force along displacement does work.Kinetic energy (KE):
KE = ½ m v2Gravitational potential energy (near Earth’s surface):
U = m g hElastic potential energy (spring):
U = ½ k x2Work–energy theorem:
Wnet = ΔKE = ½ m (vf2 − vi2)Power:
P = dW/dt (average power = W / t)Hooke’s law:
F = −k xKinematic relation (constant acceleration):
vf2 = vi2 + 2 a sNote: Use
g = 9.8 m·s−2 unless specified otherwise.Problem 1
(Work — basic)
A constant horizontal force of 25 N pulls a crate 6 m along a floor. Calculate the work done by the force.
Given:
F = 25 N, s = 6 m, angle between force and displacement θ = 0°.Work:
W = F s cos θ = 25 × 6 × cos 0° = 150 J.Answer:
150 JProblem 2
(Work with angle)
A person pulls a luggage with a force of 40 N at an angle of 30° above horizontal. If the luggage moves 10 m horizontally, find the work done by the force.
Horizontal component of force:
F_x = F cos 30° = 40 × (√3/2) ≈ 34.64 N.Work =
F_x × s = 34.64 × 10 ≈ 346.4 J.Answer: ≈
346.4 J (to 1 decimal place)Problem 3
(Work — perpendicular force)
A satellite moves in a circular orbit at constant speed. Explain the work done by the centripetal force over one complete revolution.
Centripetal force acts radially towards centre while displacement is tangential ⇒ angle 90° between force and instantaneous displacement.
Work for infinitesimal displacement:
dW = F cos 90° ds = 0. Over full revolution total work = 0.Answer:
0 J (centripetal force does no work)Problem 4
(Kinetic energy calculation)
Find the kinetic energy of a 1.5 kg ball moving at 8 m·s−1.
Given:
m = 1.5 kg, v = 8 m·s−1.KE =
½ m v2 = 0.5 × 1.5 × 82 = 0.75 × 64 = 48 J.Answer:
48 JProblem 5
(Potential energy)
A 2 kg mass is raised vertically by 3.5 m. Calculate increase in gravitational potential energy (g = 9.8 m·s−2).
Given:
m = 2 kg, h = 3.5 m, g = 9.8 m·s−2.ΔU =
m g h = 2 × 9.8 × 3.5 = 68.6 J.Answer:
68.6 JProblem 6
(Work–energy theorem)
A 4 kg cart accelerates from rest to 3 m·s−1 under a constant net force. Use work–energy theorem to find net work done on cart.
Initial speed
u = 0, final speed v = 3 m·s−1, mass m = 4 kg.ΔKE =
½ m (v2 − u2) = 0.5 × 4 × (9 − 0) = 2 × 9 = 18 J.By work–energy theorem,
Wnet = ΔKE = 18 J.Answer:
18 JProblem 7
(Friction work)
A box of mass 5 kg is dragged 10 m along floor with coefficient of kinetic friction 0.3. If pulled horizontally at constant speed, calculate work done by friction (g = 9.8 m·s−2).
Normal force
N = m g = 5 × 9.8 = 49 N.Friction force
f_k = μ N = 0.3 × 49 = 14.7 N opposite motion.Work by friction
W_f = − f_k s = −14.7 × 10 = −147 J.Answer:
−147 JProblem 8
(Spring energy)
A spring with constant
k = 500 N·m−1 is compressed by 0.04 m. Find elastic potential energy stored in spring.U =
½ k x2 = 0.5 × 500 × (0.04)2 = 250 × 0.0016 = 0.4 J.Answer:
0.4 JProblem 9
(Energy conversion)
A 0.2 kg stone is dropped from 10 m. Neglect air resistance. Find its speed just before impact and the work done by gravity. (g = 9.8 m·s−2).
Use energy conservation: initial PE = m g h = 0.2 × 9.8 × 10 = 19.6 J.
Just before impact KE = 19.6 J = ½ m v2 ⇒ v2 = (2 × 19.6) / 0.2 = 196 ⇒ v = 14 m·s−1.
Work done by gravity = gain in KE = 19.6 J.
Answer: Speed =
14 m·s−1; Work by gravity = 19.6 JProblem 10
(Work against gravity and power)
A worker lifts a 15 kg crate to a height of 2.5 m in 5 s. Calculate work done and average power (g = 9.8 m·s−2).
Work =
m g h = 15 × 9.8 × 2.5 = 367.5 J.Average power =
P = W / t = 367.5 / 5 = 73.5 W.Answer: Work =
367.5 J; Power ≈ 73.5 WProblem 11
(Work–energy with friction)
A 3 kg block is pulled 6 m along horizontal surface by 20 N horizontal force. Coefficient of kinetic friction is 0.2. Find acceleration of block assuming mass starts from rest. (g = 9.8 m·s−2).
Normal reaction
N = m g = 3×9.8 = 29.4 N; friction f_k = μ N = 0.2 × 29.4 = 5.88 N.Net force
F_net = 20 − 5.88 = 14.12 N.Acceleration
a = F_net / m = 14.12 / 3 ≈ 4.707 m·s−2.Answer:
a ≈ 4.71 m·s−2Problem 12
(Spring → KE)
A spring (k = 200 N·m−1) compressed by 0.05 m launches a 0.25 kg block on frictionless surface. Find speed of block after release.
Stored energy:
U = ½ k x2 = 0.5 × 200 × 0.052 = 100 × 0.0025 = 0.25 J.All energy → KE:
½ m v2 = 0.25 ⇒ v = √(0.5 / 0.25) = √2 ≈ 1.414 m·s−1. (Compute precisely: v = √(2U/m) = √(2×0.25 / 0.25) = √2)Answer:
v ≈ 1.414 m·s−1Problem 13
(Work & inclined plane)
A 10 kg mass is pulled up a smooth incline of vertical height 2 m. Calculate work done by gravity and by an applied upward parallel force equal to weight component (assume no friction).
Work done by gravity =
W_g = − m g h = −10 × 9.8 × 2 = −196 J (negative because gravity opposes upward motion).If pulled slowly at constant speed by force equal to weight component, applied work = +196 J (to overcome gravity), net work = 0 (no ΔKE).
Answer:
W_g = −196 J; Applied work = +196 JProblem 14
(Power & lifting)
A motor lifts 250 kg elevator up 12 m in 20 s. Find power required ignoring losses (g = 9.8 m·s−2).
Work =
m g h = 250 × 9.8 × 12 = 29,400 J.Power =
P = W / t = 29,400 / 20 = 1,470 W ≈ 1.47 kW.Answer:
1,470 W (≈ 1.47 kW)Problem 15
(Energy conservation — pendulum)
A pendulum bob of mass 0.5 kg is raised to a height of 0.2 m above its lowest point and released. Find speed at lowest point (g = 9.8 m·s−2).
Initial PE =
m g h = 0.5 × 9.8 × 0.2 = 0.98 J.At lowest point KE = 0.98 J = ½ m v2 ⇒ v2 = (2 × 0.98) / 0.5 = 3.92 ⇒ v ≈ 1.98 m·s−1.
Answer:
v ≈ 1.98 m·s−1Problem 16
(Braking work)
A car of mass 1000 kg moving at 20 m·s−1 is brought to rest by brakes. Calculate work done by brakes (change in kinetic energy).
Initial KE =
½ m v2 = 0.5 × 1000 × 400 = 200,000 J.Final KE = 0 ⇒ Work by brakes = ΔKE = 0 − 200,000 = −200,000 J (negative, removing KE).
Answer:
−200,000 J (i.e., 200 kJ of energy dissipated as heat)Problem 17
(Variable force — spring work)
How much work is required to stretch a spring with k = 150 N·m−1 from x = 0.1 m to x = 0.4 m?
Work = ∫0.10.4 k x dx = ½ k (0.42 − 0.12) = 0.5 × 150 × (0.16 − 0.01) = 75 × 0.15 = 11.25 J.
Answer:
11.25 JProblem 18
(Inclined plane with friction)
A 6 kg block is pulled 5 m up an incline making 20° with horizontal at constant speed. Coefficient of kinetic friction is 0.15. Find work done by the pulling force. (g = 9.8 m·s−2).
Vertical rise:
h = 5 sin20° ≈ 5 × 0.342 = 1.71 m.Increase in PE =
m g h = 6 × 9.8 × 1.71 ≈ 100.55 J.Normal reaction:
N = m g cos20° = 6 × 9.8 × 0.94 ≈ 55.37 N.Friction force:
f_k = μ N = 0.15 × 55.37 ≈ 8.305 N. Work against friction: W_f = f_k × s = 8.305 × 5 ≈ 41.525 J.Total work by pulling force = ΔPE + W_f ≈ 100.55 + 41.525 ≈ 142.075 J.
Answer: ≈
142.1 JProblem 19
(Energy & efficiency)
A motor does 5000 J useful work in lifting water but consumes 6500 J electrical energy. Calculate efficiency of motor.
Efficiency
η = (useful output / input) × 100% = (5000 / 6500) × 100% ≈ 76.92%.Answer: ≈
76.92%Problem 20
(Combined work & KE)
A 0.75 kg object is pushed by a net constant force and moves 4 m, increasing its speed from 2 m·s−1 to 6 m·s−1. Determine the net force.
Use work–energy theorem:
W_net = ΔKE = ½ m (vf2 − vi2).Compute ΔKE = 0.5 × 0.75 × (36 − 4) = 0.375 × 32 = 12 J.
Net work also =
F_net × s, so F_net = W_net / s = 12 / 4 = 3 N.Answer:
3 N