A swimmer at the end of a pool bends her knees, presses her feet against the wall and shoots off across the water. The wall is the only thing that pushed her forward. Yet the wall never moved, so by Lesson 1 it did no work on her. She was at rest, and now she moves fast. Where did the energy of her motion come from?
A system and its internal forces
Choose a group of bodies and call it a system. A force that one member of the system exerts on another member is an internal force. A force from anything outside the system is an external force. Two skaters pushing each other form a system. The push of A on B and the push of B on A are internal forces. Their weights and the normal forces of the ice are external.
Internal forces cancel as vectors
By Newton's third law, every internal force has a partner inside the system, equal in size and opposite in direction. So when all the internal forces are added as vectors, they cancel in pairs. The vector sum of the internal forces of any system is zero. This is why internal forces cannot push a system as a whole: a man cannot lift himself by pulling up on his own belt.
Their works need not cancel
Work depends on the force and on the displacement of the point where it acts. The two forces of a pair are equal and opposite, but the two points where they act can move by different amounts. Let the force on body A be \(\vec F\), and the force on body B be \(-\vec F\). Let their points of application move by \(\vec d_A\) and \(\vec d_B\).
\[ W_A = \vec F\cdot\vec d_A \]
Why. the work of the force on A over the displacement of its point of application.
\[ W_B = (-\vec F)\cdot\vec d_B \]
Why. by the third law, the partner force on B is \(-\vec F\). It acts over B's displacement.
\[ W_{\text{pair}} = \vec F\cdot\vec d_A - \vec F\cdot\vec d_B = \vec F\cdot(\vec d_A - \vec d_B) \]
Read it. add the two works. Only the difference of the displacements is left: the displacement of A relative to B.
So the works of a pair cancel only when \(\vec d_A = \vec d_B\), that is, when the two points of application move together. Two blocks pushed along a floor in contact move together, and the contact forces between them do no net work. The particles of a rigid body keep their distances, so its internal forces do no net work either. When the two points move apart or move closer, the pair does net work.
The internal forces of a system always add to zero as vectors. Their total work is zero only when their points of application have no relative motion along the forces. When the parts of a system move relative to one another, the internal forces can do net positive or net negative work.
Watch the readouts in Fig 4.1. Each skater moves in the direction of the push that acts on it, so each force does positive work. Neither work cancels the other; together they give \(75\,\text{J}\). The lighter skater has the larger acceleration, \(120/40 = 3\,\text{m/s}^2\) against \(120/60 = 2\,\text{m/s}^2\). In the same half second she moves further, so the same force does more work on her. The works are in the ratio \(45 : 30\), which is \(60 : 40\), the masses the other way round.
Two blocks in contact are pushed along a smooth floor by one force on the back block. What is the total work done by the two contact forces between the blocks?
Two skaters of masses \(50\,\text{kg}\) and \(75\,\text{kg}\), at rest on smooth ice, push each other apart. The push does \(60\,\text{J}\) of work on the lighter skater. How much work does it do on the heavier one, in joules?
Where the energy comes from
Before the push, both skaters were at rest. After it, they glide apart with \(75\,\text{J}\) of energy of motion between them. The external forces did no work: their weights and the normal forces are vertical, and the skaters move horizontally. So the \(75\,\text{J}\) came from inside the system. It came from chemical energy stored in the skaters' muscles, released as their arms straightened. Internal work does not create energy. It moves energy from a store inside the system into motion.
The same happens in other systems. A shell that bursts sends its pieces flying apart, and the internal forces do positive work on every piece. The store is the chemical energy of the explosive. A compressed spring between two trolleys, released, pushes them apart; the store is the energy of the compressed spring.
Pushing off a wall
Now return to the swimmer. The wall pushes her feet forward, but her feet stay on the wall while it pushes. The point of application does not move, so the wall does no work on her. Her body still gains energy of motion. That energy comes from inside her. Her leg muscles straighten, and the internal forces within her legs do positive work on the rest of her body. The wall's push is needed to change her momentum, but the energy comes from her own store. A person jumping off a floor is the same.
A girl on roller skates pushes against a wall and rolls away from it. What work does the wall's force do on her?
A shell at rest bursts into two pieces that fly apart. What is the total work done by the internal forces of the burst?
Friction between two moving parts
Internal work can also be negative. A block slides onto a long plank that rests on a smooth floor. Friction slows the block and drags the plank forward. The block moves further than the plank, so friction does more negative work on the block than positive work on the plank. By the derivation above, the total is \(-f\,s_{\text{rel}}\). Here \(f\) is the size of the friction and \(s_{\text{rel}}\) is the distance the block slides along the plank. That energy of motion becomes heat at the rubbing surfaces.
In Fig 4.2 the block slides \(8 - 2 = 6\,\text{m}\) along the plank. And \(-f\,s_{\text{rel}} = -4 \times 6 = -24\,\text{J}\), exactly the total of the two works. The friction pair is internal to the block and the plank, yet it reduced their energy of motion by \(24\,\text{J}\). That energy warmed the two surfaces. Lesson 5 checks the same \(24\,\text{J}\) from the speeds.
A block slides along a plank that is itself moving on a smooth floor. The friction between them is \(5\,\text{N}\). While the block slides \(1.2\,\text{m}\) along the plank, what is the total work done by the friction pair, in joules, with its sign?
When you jump straight up, the floor's push is the only upward force on you. Yet the floor does no work on you. Explain both parts of this.
The floor's push is an external force, and it is the only upward force, so it is what changes your momentum from zero to upward. But while it pushes, your feet stay on the floor, so its point of application does not move and it does no work. The energy of your upward motion comes from inside you. Your leg muscles straighten, and the internal forces between the parts of your body do positive work. They spend chemical energy.
When a rifle is fired, the gases push the bullet forward and the rifle backward with equal forces, for the same short time. The bullet is hundreds of times lighter, so it moves hundreds of times further while the gases push. For a \(10\,\text{g}\) bullet and a \(4\,\text{kg}\) rifle the ratio is \(400\). If the bullet leaves with \(2000\,\text{J}\), the rifle gets only \(5\,\text{J}\). That is why the shooter feels a shove, not a blow.
- Concluding that internal forces do no work because they cancel in pairs.They cancel as vectors. Their works cancel only when their points of application move together.
- Saying that the wall or the floor gives you your energy when you push off it.Its point of application does not move, so it does no work. The energy comes from your muscles, through internal work.
- Taking the work of friction between two moving bodies as \(-f\) times the distance moved over the ground.The net work of the friction pair is \(-f\) times the distance one body slides relative to the other.