Question

A diver, of mass 50 kg, climbs up to a diving
platform 1.25 m high. a) What is his weight, in N? b) What is his change in P.E.? c) Where does this energy come from? d) He walks off the platform and falls down.
What is his K.E. as he hits the water? e) What is his speed as he hits the water?

8. The same diver now climbs to the 5 m

platform, four times as high. a) What is his change in P.E. now? b) What is his speed as he hits the water? c) What do you notice about this answer?

Answer 1

Given:-

• Mass of Diver = 50kg

• Height of Platform = 1.25m

To Find:-

• What is his weight, in N

• What is his Change in P.E

• Where does this energy come from ?

• What is his K. E as he hits the water.

• What is speed as he hits the water.

Formulae used:-

• W = ma

• P. E = mgh

• K. E = ½ × m × v²

Now,

As it moves up Acceleration due to gravity acts.

So,

→ w = ma

→ w = 50 × 9.8

→ w = 490N

Hence, The weight of diver is 490N.

Now,

→ P. E = mgh

→ P. E = 50 × 9.8 × 1.25

→ P. E = 612.5J

The Change in Potential enrgy is 612.5J.

C.)

→ Potential energy come from work done against a Conservative force.

D.)

→ The K. E of diver before he hits the ground is equal to the potential energy stored in it. because the Potential energy get converted into Kinetic energy while falling down. So, K. E = 612.5J

e.)

→ mv²/2 = 612.5J

→ 50 × v² = 612.5 × 2

→ 50v² = 1225

→ v² = 1225/50

→ v² = 24.5

→ v = 4.9m/s

Therefore, The speed as he hits the water is 4.9m/s