Question

A coconut falls out of a tree 12.0 m above the ground and hits a bystander 3.00 m tall on the top of the head. It bounces back up 1.50 m before falling to the ground. If the mass of the coconut is
2.00 kg, calculate the potential energy of the coconut relative to the ground at each of the following sites:
(a) while it is still in the tree,
(b) when it hits the bystander on the head,
(c) when it bounces up to its maximum height,
(d) when it lands on the ground,
(e) when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole.

Answer 1

Answer:

A. 240 J

B. 60 J

C. 90 J

D. 0 J

E. 50 J

Explanation:

A. Determination of the potential energy of the coconut while it is still in the tree

Mass (m) = 2 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 12 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 12

PE = 240 J

B. Determination of the potential energy of the coconut when it hits the bystander on the head,

Mass (m) = 2 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 3 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 3

PE = 60 J

C. Determination of the potential energy of the coconut when it bounces up to its maximum height,

Mass (m) = 2 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 3 + 1.5 = 4.5 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 4.5

PE = 90 J

D. Determination of the potential energy of the coconut when it lands on the ground,

Mass (m) = 2 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 0 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 0

PE = 0 J

E. Determination of the potential energy of the coconut when it rolls into a ground hole, and falls 2.50 m to the bottom of the hole.

Mass (m) = 2 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 2.50 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 2.50

PE = 50 J