Question

A 0.8 kg tetherball hangs on the end of a cord. It is hit by a child and rises 2.1 m above the ground.

a. What is the maximum gravitational potential energy of the ball?

b. What was the ball’s kinetic energy at the lowest part of the swing (ground level)?

c. What was the ball’s velocity at the lowest part of the swing (ground level)?

d. Explain where the kinetic and potential energy would be exactly the same magnitude and why.

Answer 1

(a) Mass of the ball, m = 0.8 kg

Height above the ground it rises, h = 2.1 m

Maximum gravitational potential energy of the ball = m * g * h

= 0.8 *9.8 * 2.1 = 16.46 J

(b) Ball's kinetic energy at the lowest part of the swing = Maximum gravitational potential energy of the ball

= 16.46 J

(c) Expression of the kinetic energy, KE = (1/2)*m*v^2 = 16.46 J

=> 0.5*0.8*v^2 = 16.46

=> v^2 = 16.46 / (0.5*0.8)

=> v = 6.41 m/s

(d) At the middle part of the journey, means at a height of (h/2), kinetic and potential energy would be exactly the same magnitude.