Answered

A 65-kg 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 m , and at this instant he is traveling at 3.5 m/s .

Part A: What is the average force exerted by the wall on him? Express your answer with the appropriate units. FavF a v = 720 N

Part B: What is the work done by the wall on him?

Answer :

cjmejiab

Our values can be defined like this,

[tex]m = 65kg[/tex]

[tex]v = 3.5m / s[/tex]

[tex]d = 0.55m[/tex]

The problem can be solved for part A, through the Work Theorem that says the following,

[tex]W = \Delta KE[/tex]

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

[tex]\frac {1} {2} mv ^ 2[/tex]

So,

[tex]Fd = \frac {1} {2} m ^ 2[/tex]

Clearing F,

[tex]F = \frac {mv ^ 2} {2d}[/tex]

Replacing the values

[tex]F = \frac {(65) (3.5)} {2 * 0.55}[/tex]

[tex]F = 723.9N[/tex]

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

Other Questions