Work.

Step 1 of 6

A force over a distance

The fan pushes the cart forward with a steady force, and the cart moves in the direction of the push. When a force $F$⁠ acts on a body that moves a distance $d$⁠ in the direction of the force, the force does work on it:

$W = F \cdot d$⁠

The unit of work is the newton times the metre, called the joule: 1 J = 1 N·m. One joule is roughly the work you do lifting an apple by a metre.

Notice that there is no time in the formula. A force that pushes on a wall for an hour, while the wall does not move, does no work on it at all.

Step 2 of 6

Forces that do no work

Show the forces is On: see every force on the cart. The air’s push acts along the track, in the direction of motion. But the weight pulls down and the normal force pushes up: both are at right angles to the motion.

A force at right angles neither carries the cart along the track nor holds it back, so it does no work, however big it is. Only the part of a force along the motion counts.

Step 3 of 6

A force at an angle

When a force pulls at an angle $\theta$⁠ to the motion, split it into components, as you did with vectors. The component along the motion, $F\cos\theta$⁠, does the work, and the perpendicular component does nothing:

$W = F \, d \cos\theta$$W =$$F \, d \cos\theta$⁠

At $\theta = 0^\circ$⁠ this is $W = F d$⁠ again, and at $\theta = 90^\circ$⁠ the work is zero, like the weight’s on the track.

Step 4 of 6

Negative work

Now the track is rougher, and Show the forces shows the red friction arrow. Friction pulls backwards, against the motion: the angle between it and the motion is $180^\circ$⁠, and $\cos 180^\circ = -1$⁠. So its work is negative: it takes from the cart instead of giving.

Step 5 of 6

Net work and speed

Add up the work of all the forces on a body and you get the net work, which is also the work of the net force. It equals a new quantity, the body’s kinetic energy, the energy it has because it is moving:

$W = \Delta E_k, \qquad E_k = \tfrac12 m v^2$$W =$$\Delta E_k, \qquad E_k =$$\tfrac12 m v^2$⁠

Here $W$⁠ is the net work, and this is the work–energy theorem. A body that starts at rest and has net work $W$⁠ done on it moves at the speed with $\tfrac12 m v^2 = W$⁠. The speed is squared: to double the speed takes four times the work.

Try it: change the Cart mass. The same push over the same distance gives a heavy cart and a light one the same kinetic energy, but the light cart moves faster.

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Step 6 of 6

In summary

• A force that a body moves along does work $W = F d \cos\theta$⁠, in joules: 1 J = 1 N·m.
• A force at right angles to the motion, like the weight and the normal force on a level track, does no work. A force against the motion, like friction, does negative work.
• The net work equals the change in kinetic energy, $E_k = \tfrac12 m v^2$$E_k =$$\tfrac12 m v^2$⁠. In the next lesson you will see where the energy goes when a body climbs and falls.