Question

A 65-kgkg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the instant at which his fingers lose contact with the wall, his center of mass has moved 0.55 mm , and at this instant he is traveling at 4.0 m/sm/s .\

1). What is the average force exerted by the wall on him?

2). What is the work done by the wall on him?\

3). What is the change in the kinetic energy of his center of mass?

Answer 1

Solution:

From the information given,

a)

Using work energy theorem,

W = F_avgd = K.E = 1/2mv² [As he is initially at rest]

F_avg = mv²/2d

= 65*4²/2*0.55

= 945.45 N.

b) The work done by the wall is zero since there was no displacement of the wall, that is d = 0. so, W = 0 J

c) Using work energy theorem,

K.W = Fd

= 945.45 *0.55

= 520 J.

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