When is mechanical energy conserved?

Mechanical energy is the sum of an object's kinetic energy (energy of motion) and potential energy (stored energy). This total mechanical energy is conserved—meaning it stays exactly the same—only when there are no non-conservative forces doing work on the system. Think of it like moving money between a checking account and a savings account. As long as you don't spend any money outside the bank (no friction) and no one gives you extra money (no applied forces), your total bank balance remains constant, even if the amounts in the individual accounts change.

What are conservative and non-conservative forces?

To know if mechanical energy is conserved, you have to look at the forces acting on the object. Conservative forces are forces where the work done does not depend on the path taken. The main examples you will see in physics are gravity and spring forces. Non-conservative forces are things like friction, air resistance, and applied forces (like you pushing a box). Mechanical energy is only conserved if non-conservative forces do zero work.

The Conservation of Mechanical Energy Equation

When mechanical energy is conserved, the initial mechanical energy equals the final mechanical energy. We write this mathematically as Ki+Ui=Kf+UfK_i + U_i = K_f + U_f, where KK is kinetic energy and UU is potential energy. If an object falls, it loses potential energy but gains the exact same amount of kinetic energy, keeping the total sum perfectly balanced.

Where students slip up

A common mistake is assuming mechanical energy is always conserved. If a problem mentions friction, air resistance, or a completely inelastic collision, mechanical energy is lost (usually converted to heat or sound). In those cases, you must use the work-energy theorem instead, which includes the work done by non-conservative forces: Ki+Ui+Wnc=Kf+UfK_i + U_i + W_{nc} = K_f + U_f.

Worked through

A 2.0 kg ball is dropped from a height of 5.0 meters. Assuming air resistance is negligible, what is its velocity just before it hits the ground? Use g=9.8m/s2g = 9.8 m/s^2.

Because air resistance is negligible, only gravity (a conservative force) is doing work. We can use conservation of mechanical energy. Let the ground be a height of h=0h = 0. The initial energy is all potential: Ei=mghi=(2.0)(9.8)(5.0)=98JE_i = mgh_i = (2.0)(9.8)(5.0) = 98 J. The initial kinetic energy is zero because it is dropped from rest. Just before hitting the ground, the height is zero, so potential energy is zero. All the energy is now kinetic: Ef=12mvf2E_f = \frac{1}{2}mv_f^2. Setting Ei=EfE_i = E_f, we get 98=12(2.0)vf298 = \frac{1}{2}(2.0)v_f^2. Solving for vfv_f, we get 98=1.0vf298 = 1.0 v_f^2, so vf=989.9m/sv_f = \sqrt{98} \approx 9.9 m/s.

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Where this comes from: OpenStax College Physics, Chapter 7: Work, Energy, and Energy Resources · Khan Academy: Work and Energy Unit

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