work kinetic energy and power

The work done by a constant force is the scalar quantity $W = F s \cos\alpha$, where $\alpha$ is the angle between the force vector and the displacement vector, and $s$ is the magnitude of the displacement. A force directed opposite to the motion does negative work, while a force perpendicular to the displacement does zero work. If several forces act on a body, the net work equals the sum of the individual works: $W_{\text{net}} = \sum_i W_i$.

The link between work and motion is given by the work–energy theorem: $W_{\text{net}} = \Delta E_k = \frac{mv^2}{2} - \frac{mv_0^2}{2}.$ It follows from Newton's second law and is convenient because it does not require knowing the time of motion. Power describes how fast work is done: average power $P = W/t$, instantaneous power $P = F v \cos\alpha$; work and energy are measured in joules (J), power in watts (W).

A crate is pulled by a rope with a force $F = 50\ \text{N}$ at an angle $\alpha = 30^\circ$ to the horizontal and is moved $s = 4\ \text{m}$ along a horizontal floor. What work does the rope's tension force do?

A block of mass $m = 2\ \text{kg}$ slides on a horizontal surface with speed $v_0 = 5\ \text{m/s}$. The coefficient of friction is $\mu = 0.2$ and the acceleration due to gravity is $g = 10\ \text{m/s}^2$. Using the work–energy theorem, find the distance travelled before it stops.

A car of mass $m = 1200\ \text{kg}$ accelerates along a horizontal road from $v_0 = 10\ \text{m/s}$ to $v = 20\ \text{m/s}$ in $t = 5\ \text{s}$. The resistance force is constant and equal to $F_c = 500\ \text{N}$. Find: (a) the net work, (b) the work done by the engine's driving force, (c) the average engine power.

Self-check with a simulation. Use the PhET The Ramp simulation or a similar one where force, mass, and friction can be set. Do not use real heavy loads, sloped staircases, or slippery floors — that is hazardous.

1. Set $m = 2\ \text{kg}$, force $F = 10\ \text{N}$ at angle $\alpha = 0^\circ$, friction $\mu = 0$. Move the block through $s = 3\ \text{m}$ and record the travel time $t$.
2. Compute $W = F s \cos\alpha$ and compare it with the change in kinetic energy $\Delta E_k$ shown by the simulation: you should get $W \approx \Delta E_k$.
3. Repeat at $\alpha = 30^\circ$ and $60^\circ$; explain why the work decreases and becomes zero at $\alpha = 90^\circ$.
4. Turn on friction and check that $W_{\text{net}} = W_F + W_{\text{fr}} = \Delta E_k$.
5. Divide the force's work by the acceleration time to find the average power $P = W/t$ and compare it with the instantaneous power $P = F v \cos\alpha$ at the end of the motion.

To show that only the projection of the force onto the displacement determines the work, and to link the net work to the change in kinetic energy.

Diagram and graphs. Left: a block, the displacement vector $\vec{s}$ pointing to the right, and the force vector $\vec{F}$ at angle $\alpha$ to $\vec{s}$; the decomposition into $F\cos\alpha$ along $\vec{s}$ and $F\sin\alpha$ perpendicular to it is shown. Right: a bar chart of the kinetic energy $E_k$ before and after the motion, and a graph of the net work $W_{\text{net}}$ versus distance $s$ — a straight line whose slope equals $F_{\text{net}}$.

A cyclist plus bike has mass $m=70\,\mathrm{kg}$. Starting from rest, over a straight $20\,\mathrm{m}$ section, the cyclist reaches $v=8\,\mathrm{m/s}$ in $t=5.0\,\mathrm{s}$.

The work–kinetic-energy theorem says the net work equals the change in kinetic energy: $W_{\text{net}}=\Delta K=\frac12 mv^2-\frac12 m(0)^2=2240\,\mathrm{J}$. The average net force is $F_{\text{net}}=W_{\text{net}}/s=112\,\mathrm{N}$, and the average power is $P=W_{\text{net}}/t=448\,\mathrm{W}$.

A car of mass $m=1200\,\mathrm{kg}$ travels at $v_0=20\,\mathrm{m/s}$ and brakes to a full stop over $s=25\,\mathrm{m}$ in $t=3.0\,\mathrm{s}$.

The initial kinetic energy is $K_0=\frac12 mv_0^2=240000\,\mathrm{J}$. The braking force does negative work: $W=-K_0=-240000\,\mathrm{J}$, so the average braking force is $F=|W|/s=9600\,\mathrm{N}$. The average power dissipated is $P=|W|/t=80000\,\mathrm{W}=80\,\mathrm{kW}$.

A bridge crane lifts a $m=500\,\mathrm{kg}$ beam vertically through $d=12\,\mathrm{m}$ at constant speed in $t=24\,\mathrm{s}$; take $g=9.8\,\mathrm{m/s^2}$.

At constant speed the lifting force balances the weight: $F=mg=4900\,\mathrm{N}$. The lifting work is $W=Fd=58800\,\mathrm{J}$, and the power is $P=W/t=2450\,\mathrm{W}$. Because $\Delta K=0$, this work goes into a change of potential energy, linking the topic to conservation of mechanical energy.