8867 / 2027
Projectile Motion overview

Lesson 1 of 4

Weight and gravitational energy change

  • I can describe and use weight as the force experienced by a mass in a gravitational field.Syllabus

    Syllabus 8867, 5(a). Describe and use weight as the force experienced by a mass in a gravitational field.

  • I can derive gravitational potential-energy change in a uniform field from work: delta E_p = mg delta h.Syllabus

    Syllabus 8867, 5(c). Derive gravitational potential-energy change in a uniform field from work: delta E_p = mg delta h.

  • I can recall and use delta E_p = mg delta h.Syllabus

    Syllabus 8867, 5(d). Recall and use delta E_p = mg delta h.

Make a guess. You are not marked.

A ball is thrown at an angle and lands at the same height it was thrown from. Over the whole flight, what is the total work done by gravity?

  1. Negative, because gravity slowed the ball on the way up.
  2. Zero, because it ends at the same height it started from.
  3. Positive, because gravity acted on the ball the whole time.
Show the answer

Zero, because it ends at the same height it started from.

Gravity does negative work on the way up and the same amount of positive work on the way down. The net vertical displacement is zero, so the total is zero.

Weight is the gravitational force on a mass. A change in height changes the gravitational energy of the mass-Earth system; it does not change the body's mass.

Weight magnitude W = mg

Mass m is measured in kg, weight W in N, and gravitational field strength g in N/kg. Weight acts along the gravitational field. Here the field is downward and effectively uniform over the height range.

When gravity is the only significant force and mass is constant, Newton's second law gives ma = mg, so the downward acceleration has magnitude g. With upward positive, ay = -g. The units N/kg and m/s2 are equivalent, but force per unit mass and acceleration describe different quantities.

A support or scale reading is not always the weight. For a body with only upward support R and downward weight mg, Newton's second law gives R - mg = may. The reading R equals mg only when vertical acceleration is zero. An unsupported body can fall freely with R = 0 while gravitational weight remains.

Derive the gravitational energy change

Choose upward as positive and let Δh = hfinal - hinitial. The constant gravitational force is downward, opposite a positive height change. From force multiplied by displacement in its direction:

Wgravity = -mgΔh

For the work of a static gravitational field, Wgravity = -ΔEp. Combining the two statements gives:

ΔEp = mgΔh

This is the change in the mass-Earth system's gravitational potential energy. It does not assign an absolute zero at every location. Adding a common constant to the initial and final energies leaves the change unchanged.

Another way to see the result is to lift the mass slowly at constant speed with no other energy changes. The lifting force is mg, so the external agent's work is mgΔh. That positive transfer supplies the gravitational increase. If kinetic energy changes or energy dissipates, the agent's total work need not equal the gravitational increase alone.

Worked height comparison

Use the vertical rise

A 2.0 kg load is raised by a vertical height of 1.5 m in a supplied uniform field with g = 10 N/kg. Its weight is 20 N and:

ΔEp = 2.0 × 10 × 1.5 = +30 J
Wgravity = -30 J

The vertical height change sets the gravitational energy change

Two routes raise the same load by 1.5 metresA two-kilogram load starts at A. A vertical route leads to B; a longer smooth sloping route leads to C at the same final height as B. A vertical dimension marks the 1.5-metre height increase, not the length of the sloping route. With uniform downward gravitational field strength ten newtons per kilogram, the load-Earth gravitational potential-energy increase is thirty joules for either route. Blue arrows indicate movement along alternative routes, not forces. Apart from the labelled height, the route geometry is schematic.Uniform g = 10 N/kg downwardB and C have the same height.ABCVerticalLongersmooth path1.5 mSame 2.0 kg load: +30 J either way

Use the 1.5 m vertical rise, not the longer route length. This is the change in the load-Earth gravitational store. Equating it to external work also requires negligible friction and no change in kinetic energy.

The two routes have the same vertical height change. The 1.5 m bracket measures that change, not the length of a sloping route.

The gravitational increase is 30 J whether the load follows a vertical route or a longer smooth route between the same heights. On a straight smooth ramp at constant speed, a smaller parallel pulling force acts over a longer distance; the external work still supplies the same 30 J increase.

Lowering the load through the same height gives Δh = -1.5 m, so ΔEp = -30 J while gravity does +30 J work. Keep the energy-change sign separate from the sign of the named force's work.

Only the vertical displacement contributes to work by the vertical gravitational force. Do not put a sloping path length into mgΔh. Friction or a kinetic-energy change can alter the required external work without altering the gravitational change between the stated heights.

The formula assumes an effectively uniform field. It is suitable for small height changes near Earth's surface, not as one constant-g expression for all distances from Earth.

Check your understanding

A 2.0 kg load is lowered along a 3.0 m sloping path through a vertical height difference of 1.5 m. Use g = 10 N/kg. What are the gravitational potential-energy change and the work done by gravity?

  1. Potential-energy change -60 J; gravity work +60 J.
  2. Potential-energy change +30 J; gravity work -30 J.
  3. Potential-energy change -30 J; gravity work +30 J.
Show the answer

Potential-energy change -30 J; gravity work +30 J.

The load descends, so delta E_p = mg delta h = -30 J. Gravity does the opposite-signed work, +30 J. The longer sloping route does not change the result in this uniform field.