Where Does the Energy of a Stretched Catapult Come From?

In everyday language, studying is hard work, but in physics work is done only when a force moves something. That precise idea links force to energy, and energy — kinetic, potential or stored in a spring — is conserved in ways that make many hard problems simple.

This lesson covers work by constant and variable forces with the work-energy theorem, kinetic and potential energy with conservation of mechanical energy, and the elastic potential energy of a spring.

**Work done by a constant force is $W=Fs\cos \theta$, work done by a variable force is the area under its force-displacement graph, and the work-energy theorem states that the net work done on a body equals its change in kinetic energy.

Constant force:**

$$W=\vec{F}\cdot \vec{s}=Fs\cos \theta$$

- Measured in joules, where $1\text{ J}=1\text{ N m}$
- Positive when $\theta <90^{\circ }$, zero when $\theta =90^{\circ }$, and negative when $\theta >90^{\circ }$, as for friction

Worked example 1. A trolley is pulled 10 m along a floor by a 50 N force at 60° to the horizontal:

$$W=50\times 10\times \cos 60^{\circ }=250\text{ J}$$

Variable force. The work is $W={\int }_{x_1}^{x_2}F(x)dx$, the area under the force-displacement graph.

Work-energy theorem:

$$W_{\text{net}}=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$$

It follows from $v^2=u^2+2as$: multiplying by $\frac{1}{2}m$ gives $\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=mas=Fs$.

Worked example 2 — braking. A 1000 kg car at 20 m s${}^{-1}$ brakes to rest over 40 m:

$$\Delta K=0-\frac{1}{2}\times 1000\times 20^2=-2.0\times 10^5\text{ J}F=\frac{2.0\times 10^5}{40}=5000\text{ N}$$

An everyday example. Pedalling a loaded cycle rickshaw up a slope takes more work than on flat ground, because part of every push works against the component of gravity along the slope.

The substance. The work-energy theorem uses the net work of all forces — a push may do positive work while friction does negative work, and only their sum equals the change in kinetic energy.

**Kinetic energy, $\frac{1}{2}m{v}^2$, is the energy of motion, gravitational potential energy, $mgh$, is energy stored by position, and when only conservative forces such as gravity do work, their sum — the mechanical energy — stays constant.

Energy and forces:

- Kinetic energy** $K=\frac{1}{2}m{v}^2$ is always positive and linked to momentum by $K=\frac{p^2}{2m}$
- Gravitational potential energy $U=mgh$ near the Earth's surface, measured from a chosen level
- Conservative forces, such as gravity and spring forces, do work that depends only on the start and end points
- Non-conservative forces, such as friction, do path-dependent work and turn mechanical energy into heat

Conservation of mechanical energy:

$$\frac{1}{2}m{u}^2+mg{h}_1=\frac{1}{2}m{v}^2+mg{h}_2$$

Worked example 1 — a falling coconut. A 1.5 kg coconut falls 12 m from a tree, with $g=9.8$ m s${}^{-2}$:

$$v=\sqrt{2gh}=\sqrt{2\times 9.8\times 12}=15.3{\text{ m s}}^{-1}K=mgh=1.5\times 9.8\times 12=176\text{ J}$$

Worked example 2 — a playground slide. A 30 kg child starts from rest at the top of a 3.0 m slide and reaches the bottom at 6.0 m s${}^{-1}$:

$$\text{energy lost to friction}=30\times 9.8\times 3.0-\frac{1}{2}\times 30\times 6.0^2=882-540=342\text{ J}$$

An everyday example. The Bhakra Nangal dam turns the gravitational potential energy of stored water into kinetic energy as it falls, and then into electrical energy in its turbines.

The substance. Mechanical energy is conserved only when non-conservative forces do no work — with friction present, total energy is still conserved, but some mechanical energy becomes heat.

**A spring stretched or compressed by x obeys Hooke's law, $F=-kx$, and the work done against this force is stored as elastic potential energy, $U=\frac{1}{2}k{x}^2$.

Hooke's law.** Within the elastic limit, the restoring force is proportional to extension, $F=-kx$, where k is the spring constant in N m${}^{-1}$ and the minus sign shows the force opposes the displacement.

Deriving the energy. The force needed rises steadily from 0 to kx, so the work done is the area of the triangle under the force-extension graph:

$$W={\int }_0^{x}kxdx=\frac{1}{2}k{x}^2\Rightarrow U=\frac{1}{2}k{x}^2$$

Worked example 1. A spring with $k=400$ N m${}^{-1}$ is compressed by 5.0 cm:

$$U=\frac{1}{2}\times 400\times 0.050^2=0.50\text{ J}$$

Worked example 2 — launching a block. The compressed spring pushes a 0.10 kg block across a smooth floor, turning all its stored energy into kinetic energy:

$$\frac{1}{2}m{v}^2=0.50\Rightarrow v=\sqrt{\frac{2\times 0.50}{0.10}}=3.16{\text{ m s}}^{-1}$$

Worked example 3 — doubling the stretch. Compressing the same spring by 10 cm stores $\frac{1}{2}\times 400\times 0.10^2=2.0$ J — four times as much, because energy depends on $x^2$.

An everyday example. A catapult used to knock mangoes off a tree stores elastic potential energy in its stretched rubber band, and releasing it turns that energy into the kinetic energy of the stone.

The substance. Stretching and compressing by the same amount store the same energy — because of the $x^2$, the direction of x does not matter.