Question

1 a) Starting from rest, a 50 Kg person dives into water from a height of 80 m above the water surface. Air resistance is negligible. (Use g= 10 m/s²) Calculate (A) the diver’s KE just before striking the water surface (B) the velocity of the diver just before striking the water (C) the work done by gravitational force on the diver (D) the power delivered by gravitational force on the diver

b) A ball of mass m=0.2 Kg sitting at rest on the ground is tossed vertically upward with KE=90 J. Ignoring air resistance and taking g =10 m/s², calculate (A) the maximum distance ball travels from ground to top. (B) The gravitational potential energy of the ball 2 seconds after it is tossed.

c) what are the formulas used to compare initial potential energy and final kinetic energy? how do you calculate their percent differences?

Answer 1

1)

a) m=50 kg; h=80 m;

A) the potential energy is converted to the kinetic energy,

PE=mgh = 50*80*10 = 40000J

KE = PE = 40000J

B) KE = 0.5mv²

v= sqrt(2*KE / m) = sqrt(2*40000/50)

v= 40 m/s

C) Work done is change in kinetic energy

=KE= 40000J

D) using

v= u+gt

t= v/g {since u=0}\

t=40/10= 4 seconds

power = W/t = 40000/4 = 10000 Watts

b)

A) Again the KE will convert into PE

PE = KE =90 J

mgh= 90

h= 90/mg = 90 /(0.2*10)

h=45 m

B) initial velocity, u = sqrt(2K/m) =sqrt(2*90/0.2);

u= 30 m/s  

height gained in 2 seconds

using

s= ut - 0.5gt² = 30*2 - 10*0.5*2²

s=40 m