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 0.45m , and at this instant he is traveling at 3.0m/s .

A. What are the average force exerted by the wall on him?.
B. What are the work done by the wall on him?.
C. What are the change in the kinetic energy of his center of mass?

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AMB000 AMB000 · Answer 1 · 2020-02-24T12:50:28+01:00

Answer:

A) F=650N

B) W=0J

C) [tex]\Delta K=292.5J[/tex]

Explanation:

A) Considering the equation [tex]v_f^2=v_i^2+2ad[/tex], we can calculate:

[tex]F=ma=m\frac{v_f^2}{2d}=(65kg)\frac{(3m/s)^2}{2(0.45m)}=650N[/tex]

B) The work the wall does is 0J, because the fingers of the scater against it do not move. It's the muscles on his arm that do the work.

C)[tex]\Delta K=K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{(65kg)(3m/s)^2}{2}-\frac{(65kg)(0m/s)^2}{2}=292.5J[/tex]