Work, Energy and Power — Practice

Question 1

A force F = (2i + 3j) N displaces a particle by d = (4i - j) m. Find the work done.

Accepted Answer

W = F.d = (2)(4) + (3)(-1) = 8 - 3 = 5 J.

Question 2

A spring (k = 200 N/m) is compressed by 5 cm. Find the energy stored.

Accepted Answer

U = (1/2)kx^2 = (1/2)(200)(0.05)^2 = 0.25 J.

Question 3

A 1000 kg car accelerates from rest to 20 m/s in 10 s. Find the average power.

Accepted Answer

Work = delta-K = (1/2)(1000)(20^2) = 200000 J. P = W/t = 200000/10 = 20000 W = 20 kW.

Question 4

In an elastic collision a 2 kg ball at 4 m/s hits a 4 kg ball at rest. Find the final velocities.

Accepted Answer

v1' = [(m1 - m2)/(m1 + m2)]u1 = (-2/6)(4) = -4/3 m/s. v2' = [2m1/(m1 + m2)]u1 = (4/6)(4) = 8/3 m/s.

Question 5

Can kinetic energy be negative? Can potential energy be negative?

Accepted Answer

Kinetic energy K = (1/2)mv^2 is always >= 0. Potential energy is reference-dependent and can be negative, e.g. gravitational PE below the chosen reference level.

Question 6

A force varies with position as F(x) = (2x + 3) N, x in metres. Find the work done in moving a particle from x = 0 to x = 4 m.

Accepted Answer

W = integral of (2x+3) dx from 0 to 4 = [x^2 + 3x] from 0 to 4 = (16+12) - 0 = 28 J. For a variable force, work is the area under the F-x graph, computed here by direct integration rather than a single F times d.

Question 7

A particle of mass 0.5 kg moves under a potential V(x) = 2x^2 (SI units) with total energy 18 J. Find the turning points, and the maximum speed of the particle and where it occurs.

Accepted Answer

Turning points: K = 0 means V(x) = 18, so 2x^2 = 18, x = +/-3 m. Maximum speed occurs where V(x) is smallest, at x = 0, where V = 0 and all 18 J is kinetic: (1/2)(0.5)v^2 = 18 gives v^2 = 72, v = 8.49 m/s.

Question 8

A 2 kg block slides from rest down a frictionless incline of height 5 m. Find its speed at the bottom using energy conservation.

Accepted Answer

mgh = (1/2)mv^2, so v = sqrt(2gh) = sqrt(2 x 10 x 5) = sqrt(100) = 10 m/s. The mass cancels out and the angle of the incline never enters -- only the height matters for a frictionless surface.

Question 9

A 3 kg ball moving at 6 m/s collides perfectly inelastically with a stationary 2 kg ball. Find their common velocity after collision and the kinetic energy lost.

Accepted Answer

v_f = m1 v1i / (m1+m2) = (3)(6)/5 = 3.6 m/s. KE before = (1/2)(3)(36) = 54 J. KE after = (1/2)(5)(3.6)^2 = 32.4 J. KE lost = 54 - 32.4 = 21.6 J, converted to heat, sound and deformation.

Question 10

Two identical balls of mass m collide head-on; one moves at 5 m/s, the other is at rest. The coefficient of restitution is 0.8. Find their velocities after collision.

Accepted Answer

Momentum: v1' + v2' = 5 (mass cancels). Restitution: e = (v2'-v1')/(5-0) = 0.8, so v2' - v1' = 4. Solving together: v2' = 4.5 m/s, v1' = 0.5 m/s -- the first ball does not stop completely because the collision is not perfectly elastic (e < 1).

Practice — Work, Energy and Power

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