Conversion of Energy

Question

A 0.8-kg apple falls from a tree branch that is 2.5 m above the ground. How fast is it moving when it hits the ground, in m / s ? (Round your answer to one decimal place if necessary.)

Answer 1

At time 1 (t₁), the apple is hanging in the tree 2,5m above the ground. The energy it has at t₁ is gravitational potential energy and we calculate that using mass times the acceleration due to gravity times the height. So:

E₁ = mgh = (0.8)(9.8)(2.5) = 19.6 joules

At time 2 (t₂), the apple is falling from the tree and is just a micrometer off the ground (almost touching the ground). The energy it has at t₂ is kinetic energy and we calculate that using 1/2 times mass times the velocity squared. So:

E₂ = 1/2mv² = (1/2)(0.8)(v²) = 0.4v² joules

Because energy is conserved, E₁ = E₂ therefore 19.6 = 0.4v²

0.4v² = 19.6

v² = 19.6/0.4

v² = 49

v = 7.0 m/s