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Topic 7 of 20

Work & Kinetic Energy

Work and energy provide an alternative to Newton's Laws that is often faster — especially when forces vary with position or you only care about speeds at two points rather than the full trajectory. The work–energy theorem directly links the net work done on an object to its change in kinetic energy.

Key Concepts

Work

Work done by a constant force:

W = \vec{F}\cdot\vec{d} = Fd\cos\theta

, where

\theta

is the angle between force and displacement. Work is a scalar (positive, negative, or zero). Units: joules (J = N·m).

Kinetic Energy

Energy associated with motion:

K = \frac{1}{2}mv^2

. Always non-negative. Units: joules. Doubling the speed quadruples the kinetic energy.

Work–Energy Theorem

The net work done on an object by all forces equals its change in kinetic energy:

W_\text{net} = \Delta K = K_f - K_i

. This holds even when forces vary.

Work by a Variable Force

For a force that varies with position:

W = \int_{x_i}^{x_f} F(x)\,dx

, which is the area under the

F

vs.

x

graph.

Power

Rate of doing work:

P = dW/dt = \vec{F}\cdot\vec{v}

. Average power:

\bar{P} = W/\Delta t

. Units: watts (W = J/s). 1 horsepower = 746 W.

Zero Work

Work is zero when force is perpendicular to displacement (

\theta = 90°

). The normal force and centripetal force do no work since they are always perpendicular to motion.

Key Equations

Work (constant force)

W = \vec{F}\cdot\vec{d} = Fd\cos\theta

θ is the angle between the force vector and the displacement vector.

Kinetic energy

K = \tfrac{1}{2}mv^2

Energy of motion; always ≥ 0.

Work–Energy Theorem

W_\text{net} = \Delta K = K_f - K_i = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2

Relates the net work to the change in kinetic energy. Works for any net force, constant or variable.

Work by a spring

W_\text{spring} = \tfrac{1}{2}kx_i^2 - \tfrac{1}{2}kx_f^2

Work done by a spring (Hooke's Law force) as it moves from x_i to x_f from equilibrium.

Power

P = \frac{dW}{dt} = \vec{F}\cdot\vec{v}

Instantaneous power delivered by a force.

Worked Example

Finding Speed Using the Work–Energy Theorem

Problem

A 2 kg block starts from rest. A net force of 10 N acts on it over 5 m. Find its final speed.

Solution

Compute the net work done on the block:

W_\text{net} = F\,d\cos 0° = 10\times 5 = 50 \text{ J}

Apply the work–energy theorem ( $v_i = 0$ ):

W_\text{net} = \tfrac{1}{2}mv_f^2 - 0 \implies v_f = \sqrt{\frac{2W_\text{net}}{m}}

v_f = \sqrt{\frac{2\times 50}{2}} = \sqrt{50} \approx 7.07 \text{ m/s}

Answer Final speed ≈ 7.1 m/s.

Practice

Exercises

7 problems

1 of 7

A $30 \text{ N}$ force moves a box $4 \text{ m}$ in the direction of the force. How much work is done?

Your answer

2 of 7

A $10 \text{ kg}$ object moves at $6 \text{ m/s}$ . What is its kinetic energy?

Your answer

3 of 7

A net force of $25 \text{ N}$ acts over $8 \text{ m}$ on a box initially at rest. What is the box's final kinetic energy?

Your answer

4 of 7

A spring with $k = 400 \text{ N/m}$ is compressed $0.10 \text{ m}$ from its natural length. How much work was done on the spring?

Your answer

5 of 7

A machine outputs $500 \text{ W}$ of power while moving a load at $2 \text{ m/s}$ . What force does it exert?

Your answer

6 of 7

How much work does gravity do on a $3 \text{ kg}$ ball that falls $5 \text{ m}$ ? ( $g = 10 \text{ m/s}^2$ )

Your answer

7 of 7

A $2 \text{ kg}$ block starts from rest. A net force of $8 \text{ N}$ acts on it over $5 \text{ m}$ . What is its final speed?

Your answer

m/s

Key Takeaways

Work is force × displacement × cosine of the angle between them. Perpendicular forces do no work.
The work–energy theorem ( $W_\text{net} = \Delta K$ ) is a scalar equation — often faster than Newton's Second Law when you only need speeds.
Friction does negative work, reducing kinetic energy.
Power is the rate of work; $P = \vec{F}\cdot\vec{v}$ is useful when force and velocity are known at an instant.
Work done by a spring is $\frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$ — it can be positive or negative depending on direction of compression/extension.