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Conservation of mechanical energy
Gravitational potential energy near Earth's surface (ΔU = mgΔh), conservation of mechanical energy for falling objects, slides, pendulums and roller coasters, energy losses to friction, and choosing between the energy and force methods.
In the Snowy Mountains, Tumut 3 power station can pump water uphill into a high reservoir when electricity is cheap and plentiful, then let it run back down through turbines when demand peaks. The water is a giant rechargeable battery. Its stored energy is simply its weight times how high it has been lifted.
When you lift something, the work you do isn't lost. It is stored as gravitational potential energy, ready to turn back into kinetic energy when the object falls. If nothing else takes energy away, the total of kinetic and potential energy stays the same. That one idea solves a lot of problems that would be messy with forces.
Gravitational potential energy
Lifting a mass $m$ at constant speed through a height $\Delta h$ takes an upward force $mg$ acting over $\Delta h$, so the work done is $mg\Delta h$. That work is stored as a change in gravitational potential energy:
$$\Delta U = mg\Delta h$$This formula assumes $g$ is constant, which is true to within 1% for heights up to about 30 km above Earth's surface. Only changes in potential energy matter, so you can choose any convenient reference level (usually the lowest point in the problem) and call $U = 0$ there.
A 55 kg student climbs from the ground floor to the third floor of a school, a vertical height of 10.5 m. Find the gain in gravitational potential energy.
Conservation of mechanical energy
Mechanical energy is the sum of kinetic and potential energy. If the only force doing work is gravity (no friction, no air resistance, no engine), mechanical energy is conserved:
$$K_i + U_i = K_f + U_f \quad\text{or}\quad \tfrac12 mu^2 + mgh_i = \tfrac12 mv^2 + mgh_f$$Forces that act perpendicular to the motion, like the normal force from a smooth track or the tension in a pendulum string, do no work, so they don't change the mechanical energy. This is why energy methods are so powerful on curved tracks: you don't need to know the shape, only the heights.
A 30 kg child starts from rest at the top of a slide 3.0 m high. Ignoring friction, how fast is the child moving at the bottom?
A pendulum bob swings through its lowest point at 2.4 m s⁻¹. How high above the lowest point does it rise?
When friction acts
Friction and air resistance do negative work, turning mechanical energy into thermal energy (and some sound). Total energy is still conserved, but mechanical energy isn't:
$$K_i + U_i = K_f + U_f + Q, \qquad Q = f\,d$$where $f$ is the friction force and $d$ is the distance slid.
The 3.0 m slide is a straight ramp at 30° to the horizontal, and the coefficient of kinetic friction between the child's clothes and the slide is 0.20. Find the speed at the bottom.
Energy or forces?
Both methods give the same answer, but each has its strengths:
- Use energy when you need a speed at a particular height or position and don't care about time, especially on curved paths.
- Use forces ($F = ma$ plus the equations of motion) when you need time or acceleration, or when the forces are given directly.
Check yourself on the frictionless slide using forces: $a = g\sin30° = 4.9$ m s⁻² over $L = 6.0$ m gives $v = \sqrt{2(4.9)(6.0)} = 7.7$ m s⁻¹, the same as the energy method.
Storing energy with height
Pumped hydro stores energy as gravitational potential energy on a huge scale. Snowy 2.0, under construction in the Kosciuszko National Park, will link two existing reservoirs about 700 m apart in height through 27 km of tunnels. It is designed to supply 2,200 MW for about a week, roughly 350 GWh, making it one of the largest energy-storage projects in the world. Each cubic metre of water (1000 kg) falling 700 m releases $1000 \times 9.8 \times 700 \approx 6.9$ MJ.
- With no friction and $h$ = 3.0 m, press Slide. The child reaches 7.67 m/s, and the bars show 882 J moving from potential to kinetic.
- Change the angle or the child's mass with no friction: the speed at the bottom stays 7.67 m/s.
- Set μk = 0.20 at 30°: 306 J becomes heat and the child reaches 6.20 m/s. Now try 45°: a steeper slide is shorter and wastes less.
- Set μk = 0.50 at 25°. Since 0.50 is more than tan 25° = 0.47, the child stays put.
Test yourself
A 0.50 kg ball is dropped from 2.0 m. Ignoring air resistance, its speed just before landing is
$v = \sqrt{2gh} = \sqrt{39.2} = 6.26$ m s⁻¹.
Two balls are thrown from the same cliff at the same speed, one horizontally and one straight up. Ignoring air resistance, when they reach the ground below
Both start with the same $K$ and $U$ and lose the same $U$, so they finish with the same $K$ and therefore the same speed, although in different directions.
A roller coaster car starts from rest at 40 m and passes over a hill 25 m high. Ignoring friction, its speed at the top of the hill is
It has dropped 15 m: $v = \sqrt{2 \times 9.8 \times 15} = 17.1$ m s⁻¹.
Which force does no work on a child swinging on a playground swing?
The chains pull along the radius, perpendicular to the child's motion.
Label the energy at each point on the track.
Is mechanical energy conserved in each case? (Ignore air resistance unless it is mentioned.)
The change in gravitational potential energy near Earth's surface is $mg$ times the change in vertical . If only gravity does work, the sum of kinetic and potential energy, called energy, is conserved. Friction transforms mechanical energy into energy. The speed of an object sliding down a frictionless ramp depends only on the it drops, not on the angle.
A 70 kg skier starts from rest at the top of a straight slope 120 m long, inclined at 20°. The coefficient of kinetic friction is 0.08 and air resistance can be ignored. (a) Use energy methods to calculate the skier's speed at the bottom. (4 marks) (b) Explain why the speed would be greater on a steeper slope of the same vertical height. (2 marks)
(a) Height: $h = 120\sin20° = 41.0$ m. Potential energy lost: $U = 70 \times 9.8 \times 41.0 = 2.82 \times 10^4$ J.
Friction: $f = \mu_k mg\cos20° = 0.08 \times 70 \times 9.8 \times 0.940 = 51.6$ N. Thermal energy: $Q = 51.6 \times 120 = 6.19 \times 10^3$ J.
Kinetic energy at the bottom: $K = 2.82 \times 10^4 - 6.19 \times 10^3 = 2.20 \times 10^4$ J, so $v = \sqrt{\dfrac{2 \times 2.20 \times 10^4}{70}} = 25$ m s⁻¹.
(b) The potential energy released depends only on the height, so it is the same. On a steeper slope the path is shorter, and the normal force $mg\cos\theta$ is smaller, so the friction force is smaller too. Both reduce the work done against friction, $Q = \mu_k mg\cos\theta \times L$, so more of the potential energy becomes kinetic energy and the skier is faster at the bottom.