Work is the energy transferred when a force acts over a distance.
The basic formula is \$W = F\,s\$ (force times displacement).
⚡️ Think of pushing a shopping cart: the harder you push (larger \$F\$) and the farther you push (larger \$s\$), the more work you do.
When you see a question about potential energy change, remember to start from the work done by the force.
Write down \$W = F\,s\$ and then replace \$F\$ with the appropriate force (e.g., \$mg\$ for gravity).
This shortcut often saves time in the exam.
In a uniform field near Earth’s surface, the weight of an object is constant:
\$F = mg\$, where \$m\$ is mass and \$g \approx 9.81\,\text{m/s}^2\$.
Choose the direction of displacement.
Let the object move upward by a distance Δh. The displacement vector is \$s = \Delta h\$ (positive upward).
Write the work done by gravity.
Gravity acts downward, opposite to the upward displacement, so the work done by gravity is negative:
\$W_{\text{gravity}} = -mg\,\Delta h\$.
Relate work to potential energy change.
By definition, the change in gravitational potential energy is the negative of the work done by gravity:
\$ΔEP = -W{\text{gravity}}\$.
Substitute.
Plugging in the expression for \$W_{\text{gravity}}\$ gives:
\$ΔE_P = -(-mg\,\Delta h) = mg\,\Delta h.\$
So, when an object rises by Δh in a uniform field, its gravitational potential energy increases by \$mgΔh\$.
If it falls, Δh is negative and the potential energy decreases accordingly.
A 5 kg box is lifted 3 m above the ground.
ΔEP = \$5\,\text{kg} \times 9.81\,\text{m/s}^2 \times 3\,\text{m} = 147.15\,\text{J}\$ (≈147 J).
The box now has 147 J more gravitational potential energy than before.
Kinetic energy (KE) is the energy of motion:
\$KE = \tfrac{1}{2}mv^2\$.
When an object falls, its potential energy converts to kinetic energy (ignoring air resistance).
Energy conservation: \$ΔE_P + ΔKE = 0\$.
For questions involving both potential and kinetic energy, write the energy conservation equation:
\$E{\text{initial}} = E{\text{final}}\$.
Then substitute \$E = KE + EP\$ and solve for the unknown quantity.
| Quantity | Formula | Units |
|---|---|---|
| Work | \$W = F\,s\$ | J |
| Gravitational Potential Energy Change | \$ΔE_P = mg\,Δh\$ | J |
| Kinetic Energy | \$KE = \tfrac{1}{2}mv^2\$ | J |