A 26 kg diver initially at rest, jumps into the water from a height of 8.5 m. 1.

Question

A 26 kg diver initially at rest, jumps into the water from a height of 8.5 m.

1. What is the work done by gravity on the diver (use g=9.8 m/s2)?

2. What is the speed of the diver at that distance?

3. If the same person jumps while attached to an initially un-stretched bungee cord (spring) with a spring constant 30 N/m, what would be the distance the diver falls towards the ground before coming to a complete stop?

4. What would be the speed of the diver attached to the bungee cord (spring) when s(h)e fell 8.5 m towards the ground?

5. Is the speed of the diver falling the same distance 8.5 m different when attached to the spring compared to the freely falling diver?

A) The speed of the object attached to the spring is the same as in the free fall because work done by the gravity is the same and depends only on the distance of the fall.

B) The speed of the object attached to the spring is larger because the spring provides a steadier motion and the losses of energy due to air friction are smaller.

C) The speed of the object attached to the spring is smaller because gravity does smaller amount of work when object is attached to the spring.

D) The speed of the object attached to the spring is the same as in the free fall because gravity is balanced by the elastic force of the spring at all times.

E) The speed of the object attached to the spring is larger because its kinetic energy has contributions both from the work done by gravity and by the spring.

F) The speed of the object attached to the spring is smaller because work done by gravity is distributed into kinetic energy of the object and potential energy of the spring.

Explanation / Answer

here,

2)

mass , m = 26 kg

height , h = 8.5 m

a)

the work done by gravity on the diver , W = m * g * h

W = 26 * 9.81 * 8.5 J

W = 2168.01 J

b)

let the speed of the diver at that distance be v

using conservation of energy

kinetic energy gained= potential eenrgy lost

0.5 * m* v^2 = 2168.01

0.5 * 26 * v^2 = 2168.01

solving for v

v = 12.9 m/s

the speed of the diver at that distance is 12.9 m/s

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