5.1, 5.2 - Work, kinetic energy and potential energy

Work done by a force

Would a heavy stone reach the ground faster than a light stone if both fell from rest through the same height in a vacuum? The heavy stone loses more gravitational potential energy. Yet that does not give it a greater final speed. To see why, start with what a force transfers when its point of application moves.

A force does work when there is displacement in its direction. Pulling a box transfers energy to it; friction can transfer energy from its motion to the internal energy of the box and floor. Work measures an energy transfer, rather than something contained in the force.

If a constant force acts at an angle to a straight displacement, resolve it along the displacement. Its perpendicular component does no work on that motion.

A force in the direction of motion does positive work. A force opposite to motion does negative work: $\cos180^{\circ}=-1$. If you are asked for the work done against friction, report the positive energy that must be supplied to overcome it.

[DIAGRAM: asset_name: The part of a force that does work; asset_slug: energy-work-components; caption: Only the component parallel to the displacement contributes to this work; description: A box is pulled by a force F at 30 degrees above the horizontal. A horizontal arrow from the box shows displacement s to the right. A dashed horizontal force component is labelled F cos theta. The force arrow and its component begin at the same point on the box.]

A force is not enough on its own: a wall does not gain mechanical energy just because you push it without moving it. Your muscles may still transfer chemical energy internally, which is a different process.

Changes in gravitational potential energy

Raising a mass separates it from Earth against the gravitational force. The work you supply can increase the gravitational potential energy of the Earth–object system. Near Earth's surface, we model the field as uniform, so the weight has constant magnitude $mg$.

Imagine lifting a mass slowly at constant speed through a vertical height $\Delta h$. The upward lifting force equals its weight. Its work is therefore

There is no change in kinetic energy, so this work gives the increase in gravitational potential energy. This derivation uses the weight's constant value, not the total length of a sloping path.

Only a change in height matters here. Choose a convenient reference height and use it consistently. A descent has negative $\Delta h$, so the gravitational potential energy decreases; gravity then does positive work.

Check which displacement controls the energy change.

Kinetic energy and net work

Speeding up an object requires a transfer of energy to its motion. The faster it is already moving, the more energy is needed for the same further increase in speed. This is why kinetic energy is not simply proportional to speed.

You can derive its expression from mechanics you already know. Consider a mass $m$ accelerating in a straight line from speed $u$ to speed $v$, under a constant resultant force $F$. With displacement $s$, Newton's second law and the constant-acceleration equation give

For an object accelerated from rest, the net work transferred to its motion is $\tfrac12mv^{2}$. This motivates the definition of kinetic energy. The derivation uses a constant force; the resulting work–energy relationship also applies when the force changes.

Doubling the speed quadruples the kinetic energy. That is a real physical consequence of the square, not a calculator trick: stopping from twice the speed requires four times the energy transfer. With the same constant braking force, the stopping distance is four times as large.

The net work is the sum of the work by all forces. If a motor transfers $1000\,\mathrm{J}$ to a trolley while friction removes $300\,\mathrm{J}$ from its motion, its kinetic energy increases by $700\,\mathrm{J}$.

Conservation of energy

An energy account must include every relevant transfer. Energy can change form or move between objects, but the total does not disappear when an object slows down.

In a fall with negligible air resistance, the loss of gravitational potential energy equals the gain in kinetic energy. For a stone falling from rest through height $h$ in a uniform field,

The mass cancels because a larger mass loses proportionally more potential energy but also needs proportionally more energy for any given speed. That answers the opening puzzle: the heavier stone has more kinetic energy, but the same speed after the same drop.

Real falls can transfer energy to the surroundings through air resistance. Mechanical energy, meaning kinetic plus potential energy here, then decreases while total energy remains conserved. Avoid saying that friction “uses up energy”. It transfers energy to internal energy and sometimes sound.

Now explain the physics without relying on identical wording.

Quick Recap

• Work is force multiplied by displacement in the force's direction; opposite forces can do negative work.
• In a uniform field, a vertical rise changes gravitational potential energy by $mg\Delta h$.
• Kinetic energy is $\tfrac12mv^{2}$; net work changes it by $\tfrac12m(v^{2}-u^{2})$.
• Total energy is conserved, including transfers to internal energy when resistance acts.

In the stated vacuum fall, the stones land together: the heavier stone gains more kinetic energy but reaches the same speed because mass cancels from $mgh=\tfrac12mv^{2}$.