Help Gr11 Physics

Question

Figure 1.5 shows a diver on a 10-metre high diving board above a swimming pool

A. The diver has a weight of 700N and climbs up to the diving board in 7.0 seconds

I) Calculate the useful work done by the diver in climbing to the diving board.

II) State how much the gravitational potential energy of the diver increases on climbing to the diving board

III) On an average state how much useful energy does the diver gain per second in climbing to the diving board.

IV) Calculate the power of the diver in climbing up to the board.

B. The diver steps off the diving board and falls towards the water. At the instant, before the diver hits the water

I) Calculate by how much the diver's potential energy will decrease

II)Calculate by how much the diver's kinetic energy will increase

please provide explanations also. I got transferred in this class against my own free will.

Answer 1

A I) Work = force * displacement, the force is the weight mg in this case, and W = (700 N) (10 m) = 7,000 J.

II) This is the same as the change in GPE, mgh = +7,000 J.

III) In 1 second, the average energy gain is (7,000 J) / (7.0 s) = 1,000 J/s.

IV) Power is work/time = 1,000 J/s.

B I) GPE will decrease to zero as height decreases, so change = GPE_f - GPE_i = 0 - 7,000 = -7,000 J.

II) By conservation of mechanical energy (GPE + KE), Ke will increase at the same rate as GPE decreases, KE_f - KE_i = 7,000 - 0 = +7000 J.

BTW, from this you can calculate the speed of the diver just before hitting the water: KE_f = 1/2 m v², mg is given as 700 N, so mass m = 70 kg for g = 10 m/s². Then 7,000 J = (70 kg /2) v², v² = (7,000 kg m²/s²) / (35 kg) = 200 m² /s², v = 14 m/s.