Part 1 · Chapter 07

Kinetic Energy and Work

A new way to solve motion problems — through energy transferred by forces

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 40 min
i What you'll learn
  • That kinetic energy \(K = \tfrac{1}{2}mv^{2}\) is the energy of motion, measured in joules.
  • That work is energy transferred to or from a body by a force — and how to compute it for a constant force.
  • The work–kinetic energy theorem: the net work done equals the change in kinetic energy.
  • How to find the work done by gravity, by a spring (Hooke's law), and by a variable force (an integral).
  • What power means — the rate of doing work — and how to compute it from force and velocity.
Section 7-1

What Is Physics?

A central goal of physics is to understand something everyone talks about: energy. Flying across an ocean, lifting cargo to an orbiting station, throwing a fastball — all require it, and our entire civilization runs on acquiring and using it. This chapter introduces a powerful, complementary way to attack motion problems — not through forces and acceleration, but through energy and work.

Section 7-2

What Is Energy?

Energy is a scalar number we associate with the state of one or more objects. Change an object — make it move, say — and the number changes. The reason energy is so useful rests on one deep property of our universe:

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The principle of energy conservation
Energy can be transformed from one type to another and transferred between objects, but the total amount never changes.

Think of energy like money across many bank accounts: you can move it around — even electronically, with nothing material flowing — but the grand total is always conserved. No exception has ever been found. This chapter studies just one type (kinetic energy) and one way to transfer it (work).

Section 7-3

Kinetic Energy

Kinetic energy is the energy of motion — the faster a body moves, the more it has; at rest it is zero. For a mass \(m\) at speed \(v\) well below light speed:

Kinetic energy
\[ K = \tfrac{1}{2} m v^{2} \]
The SI unit is the joule (J): 1 J = 1 kg·m²/s². A 3.0 kg duck flying at 2.0 m/s carries 6.0 J.
Section 7-4

Work

Speed a body up or slow it down with a force, and you change its kinetic energy — you have transferred energy to or from it. That transfer, accomplished by a force, is work.

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Definition of work
Work is energy transferred to or from an object by means of a force.

Energy transferred to the object is positive work; energy taken from it is negative work. Work is a scalar with the same unit as energy. Note: this is not the everyday "work" — pushing uselessly on a wall tires you but does zero work on the wall.

Section 7-5

Work and Kinetic Energy

For a constant force \(F\) acting through a displacement \(d\) at angle \(\varphi\) between them, only the component along the displacement does work:

Work done by a constant force
\[ W = F d \cos \varphi = F \cdot d \]
The perpendicular component does zero work. φ < 90° gives positive work; φ > 90° gives negative work; φ = 90° gives none. For several forces, the net work is the sum of the individual works.
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The work–kinetic energy theorem
The net work done on a particle equals the change in its kinetic energy.

\(\Delta K = K_{f} - K_{i} = W\), or equivalently \(K_{f} = K_{i} + W\). Positive net work speeds a particle up; negative net work slows it down. If \(K = 5 J\) and a net 2 J is transferred in, the final K is 7 J; transfer 2 J out and it is 3 J.

Section 7-6

Work Done by the Gravitational Force

For an object moving through a displacement \(d\), gravity (magnitude \(mg\)) does work \(W_{g} = mgd \cos \varphi\). A rising object has \(\varphi = 180^{\circ}\), so gravity does \(-mgd\) (removing kinetic energy); a falling object has \(\varphi = 0^{\circ}\), so gravity does \(+mgd\) (adding it).

Lifting and lowering (object at rest before and after)
\[ W_{a} = -W_{g} \qquad (\text{when} K_{f} = K_{i}) \]
If a body starts and ends at rest, your applied force transfers exactly as much energy to it as gravity transfers from it — independent of how the force varies during the lift.
Section 7-7

Work Done by a Spring Force

A spring pushes or pulls its free end back toward the relaxed position — a restoring force proportional to the stretch or compression. This is Hooke's law:

Hooke's law
\[ F_{x} = -k x \]
k is the spring constant (N/m), a measure of stiffness; the minus sign means the force opposes the displacement. The spring force is a variable force — it changes with x.

Because the force varies, we integrate to get the work it does as the free end moves from \(x_{i}\) to \(x_{f}\):

Work done by a spring
\[ W_{s} = \tfrac{1}{2} k x_{i}^{2} - \tfrac{1}{2} k x_{f}^{2} \qquad (\text{and} W_{s} = -\tfrac{1}{2} k x^{2} \text{if} x_{i} = 0) \]
Ws is positive if the block ends nearer the relaxed position, negative if it ends farther away, zero if it ends the same distance from x = 0.
Section 7-8

Work Done by a General Variable Force

When a force changes with position, we slice the path into tiny segments, treat the force as constant on each, and sum. In the limit the sum becomes an integral — and geometrically, the work is the area under the F(x) curve:

Work done by a variable force
\[ W = \int _{x_{i}}^{x_{f}} F(x) dx \]
In three dimensions, integrate each component: W = ∫Fxdx + ∫Fydy + ∫Fzdz. Carrying the integral through with Newton's second law reproduces the work–kinetic energy theorem exactly — it holds for variable forces too.
Section 7-9

Power

Power is how fast a force does work. Over a time interval it is the average; at an instant it is the derivative:

Average and instantaneous power
\[ P_{\text{avg}} = W / \Delta t \qquad P = dW/dt \qquad P = F v \cos \varphi = F \cdot v \]
The SI unit is the watt (W): 1 W = 1 J/s. Also 1 hp = 746 W, and 1 kW·h = 3.60 MJ. The last form lets you get power directly from a force and the velocity it acts along.
Sign carries over to power. A force with a component along the velocity supplies energy (positive power); one opposing the velocity removes it (negative power). A purely centripetal force — always perpendicular to the velocity — delivers zero power, which is why a whirling string does no work on the mass.
Worked Examples

Putting It to Work

1 A staged train crash — kinetic energy

Problem. In an 1896 publicity stunt, two locomotives (each weighing \(1.2\times 10^{6} N\)) were run head-on from rest along a 6.4 km track with constant acceleration \(0.26 m/s^{2}\). Find the total kinetic energy just before impact.

Solution. Each travels half the track, \(3.2\times 10^{3} m\). From \(v^{2} = 2a(x - x_{0})\), \(v = \sqrt{2\cdot 0.26\cdot 3200} \approx 40.8 m/s\). Each mass is \(m = (1.2\times 10^{6})/9.8 \approx 1.22\times 10^{5} \mathrm{kg}\).

Total kinetic energy of both
\[ K = 2(\tfrac{1}{2} m v^{2}) = (1.22\times 10^{5})(40.8)^{2} \approx 2.0\times 10^{8} J \]

Two hundred million joules — "like an exploding bomb," as the era's reporters put it.

2 Two spies sliding a safe — work and the W–K theorem

Problem. Two thieves slide a 225 kg safe \(8.50 m\) across a frictionless floor. Spy 001 pushes 12.0 N at 30.0° below horizontal; spy 002 pulls 10.0 N at 40.0° above horizontal. Find the net work, the work by gravity and the normal force, and the final speed (starting from rest).

Work by each force, then the speed
\[\begin{gathered} W_{1} = (12.0)(8.50)\cos 30.0^{\circ} \approx 88.3 J \qquad W_{2} = (10.0)(8.50)\cos 40.0^{\circ} \approx 65.1 J \\ W = W_{1} + W_{2} \approx 153 J \qquad W_{g} = W_{N} = 0 (\varphi = 90^{\circ}) \\ v_{f} = \sqrt{2W/m} = \sqrt{2\cdot 153/225} \approx 1.17 m/s \end{gathered}\]

Gravity and the normal force act perpendicular to the motion, so they transfer no energy. The 153 J of work becomes 153 J of kinetic energy, giving the safe a speed of about 1.17 m/s.

3 An accelerating elevator cab — gravity, tension, net work

Problem. A 500 kg cab descending at \(4.0 m/s\) has its cable slip, so it falls \(d = 12 m\) with acceleration \(a = g/5\) downward. Find the work by gravity, the work by the cable's pull, the net work, and the final kinetic energy.

Solution. Gravity acts along the motion (\(\varphi = 0^{\circ}\)); the cable's tension \(T = m(g - a) = (4/5)mg\) acts opposite it (\(\varphi = 180^{\circ}\)).

Each work, then the kinetic energy
\[\begin{gathered} W_{g} = mgd = (500)(9.8)(12) \approx 59 \mathrm{kJ} \\ W_{T} = -(4/5)mgd \approx -47 \mathrm{kJ} \\ W = W_{g} + W_{T} \approx 12 \mathrm{kJ} \qquad K_{f} = \tfrac{1}{2}mv_{i}^{2} + W \approx 4.0 \mathrm{kJ} + 12 \mathrm{kJ} \approx 16 \mathrm{kJ} \end{gathered}\]

Note \(W_{T}\) is not simply \(-W_{g}\) here — because the cab accelerates, its kinetic energy changes, so the lifting/lowering shortcut doesn't apply.

4 A canister compressing a spring — spring work

Problem. A 0.40 kg canister slides at \(0.50 m/s\) on a frictionless counter into a spring of constant \(k = 750 N/m\). By how much is the spring compressed when the canister is momentarily stopped?

Solution. The spring's work is \(W_{s} = -\tfrac{1}{2}kd^{2}\), and by the work–kinetic energy theorem this equals \(K_{f} - K_{i} = 0 - \tfrac{1}{2}mv^{2}\):

Set spring work equal to the lost kinetic energy
\[ 0 - \tfrac{1}{2} m v^{2} = -\tfrac{1}{2} k d^{2} \qquad \;\Longrightarrow\; \qquad d = v \sqrt{m/k} = (0.50)\sqrt{0.40/750} \approx 1.2\times 10^{-2} m = 1.2 \mathrm{cm} \]

All of the canister's kinetic energy goes temporarily into the spring, which compresses 1.2 cm before pushing it back.

5 Two forces on a sliding box — power

Problem. A box slides right at \(v = 3.0 m/s\) on a frictionless floor. Force \(F_{1} = 2.0 N\) points backward (180° to v); force \(F_{2} = 4.0 N\) points 60° above the floor. Find each power and the net power.

Instantaneous power, P = F v cos φ
\[\begin{gathered} P_{1} = (2.0)(3.0)\cos 180^{\circ} = -6.0 W \\ P_{2} = (4.0)(3.0)\cos 60^{\circ} = +6.0 W \\ P_{\text{net}} = P_{1} + P_{2} = 0 \end{gathered}\]

One force removes energy at 6.0 W while the other supplies it at 6.0 W, so the net rate of energy transfer is zero. The kinetic energy — and the speed — hold steady at 3.0 m/s.

Review

Chapter Summary

✓ Kinetic energy

\(K = \tfrac{1}{2}mv^{2}\), the energy of motion, measured in joules (1 J = 1 kg·m²/s²).

✓ Work

Energy transferred by a force. Constant force: \(W = Fd \cos \varphi = F\cdot d\). Only the component along the displacement counts.

✓ W–K theorem

\(\Delta K = K_{f} - K_{i} = W_{\text{net}}\). Positive net work speeds up; negative slows down.

✓ Gravity & springs

\(W_{g} = mgd \cos \varphi\); Hooke's law \(F_{x} = -kx\) gives \(W_{s} = \tfrac{1}{2}kx_{i}^{2} - \tfrac{1}{2}kx_{f}^{2}\).

✓ Variable force

\(W = \int F(x)dx\) — the area under the F(x) curve.

✓ Power

Rate of doing work: \(P = dW/dt = F\cdot v\), in watts (1 W = 1 J/s).

Practice

Problems

Energy methods often beat force-and-acceleration methods: if you only need a speed at the end of a displacement, the work–kinetic energy theorem skips the time entirely. Take \(g = 9.8 m/s^{2}\).

  1. A 2.9×10⁵ kg Saturn V rocket reaches 11.2 km/s. What is its kinetic energy?
  2. The only force on a 2.0 kg canister has magnitude 5.0 N. Its velocity changes from 4.0 m/s along +x to 6.0 m/s along +y. How much work did the force do?
  3. A 12.0 N force of fixed orientation acts as a particle moves through \(d = (2.00\hat{\imath} - 4.00\hat{\jmath} + 3.00\hat{k}) m\). Find the angle between force and displacement if the kinetic energy changes by (a) +30.0 J and (b) −30.0 J.
  4. A helicopter lifts a 72 kg astronaut 15 m on a cable with acceleration g/10. Find the work done on her by (a) the cable and (b) gravity, and (c) her kinetic energy and (d) speed just before reaching the helicopter.
  5. A cord lowers a block of mass M a distance d at constant downward acceleration g/4. Find (a) the work by the cord, (b) the work by gravity, (c) the block's kinetic energy, and (d) its speed.
  6. We must apply 360 N to hold a spring-block at x = 4.0 cm. We then pull the block to x = 11 cm and release it. How much work does the spring do as the block moves from xᵢ = 11 cm to (a) x = 3.0 cm and (b) x = −5.0 cm?
  7. The only force on a 2.0 kg body is \(F_{x} = -6x N\) (x in metres). The velocity at x = 3.0 m is 8.0 m/s. Find (a) the velocity at x = 4.0 m and (b) the positive x where the speed is 5.0 m/s.
  8. A 5.0 kg block on a frictionless surface is acted on by a force that varies with position; the area under its F(x) graph from x = 0 to x = 8.0 m is to be found by integration. (Sketch a plausible F(x) and compute the work.)
  9. A 100 kg block is pulled at a steady 5.0 m/s across a floor by a 122 N force angled 37° above horizontal. At what rate does the force do work?
  10. A ski lift raises 100 passengers averaging 660 N each to a height of 150 m in 60.0 s at constant speed. What average power is required of the lifting force?
Tip: forces perpendicular to the motion — gravity on a horizontal slide, the normal force, a centripetal pull — do zero work. Spotting these before you compute anything often collapses a messy problem to a single term.