Question

A 3.2 kg sloth hangs 3.0 m above the ground. (a) What is the gravitational potential energy of the sloth – Earth system if we take the reference point y = 0 to be at the ground? If the sloth drops to the ground and air drag on it is assumed to be negligible, what are the (b) kinetic energy and (c) speed of the sloth just before it reaches the ground?

Answer 1

Hint
The gravitational potential at a point in a gravitational field is the work done per unit mass that would have to be done by some externally applied force to bring a massive object to that point from some defined position of zero potential, usually infinity . The two definitions are equivalent.

Complete step by step answer
According to the question given that,
The mass of the sloth m = 3.2 Kg
And the sloth is hanged at point h = 3 m above the ground,
A ) Let, the gravitational potential energy ∆U can be calculated from the height h,
According to the formula,
$\Delta U = mgh = 3.2 \times 3 \times 9.8 = 94.08J$ [ assume the value of g = 9.8 m/s2]
B ) As the air drag is negligible, the change in kinetic energy ∆K.E is equal to the change in potential energy ∆U. As the sloth stops, so, its kinetic energy $k_h$ is zero.
$\Delta K.E = \Delta U = {K_0} - {K_h}$
$\therefore {K_0} - 0 = U = 94.08J$ [Where $K_0$ is the kinetic energy just before reaching the ground]
C ) According to the formula,
$0 - \dfrac{1}{2}mv_0^2 = mgh$
$v_0$ is the speed of the sloth just before reaching the ground,
$\Rightarrow \dfrac{1}{2}v_0^2 = gh$
$\Rightarrow {v_0} = \sqrt {(2 \times 3.2 \times 3)}$
$\Rightarrow \therefore {v_0} = 7.67m/s$.

Note
The kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.