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nasty-shy [4]
2 years ago
8

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?
Physics
1 answer:
Marrrta [24]2 years ago
6 0

Our values can be defined like this,

m = 65kg

v = 3.5m / s

d = 0.55m

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

W = \Delta KE

Where

KE = Kinetic energy,

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

W = Fd

and kinetic energy by

\frac {1} {2} mv ^ 2

So,

Fd = \frac {1} {2} m ^ 2

Clearing F,

F = \frac {mv ^ 2} {2d}

Replacing the values

F = \frac {(65) (3.5)} {2 * 0.55}

F = 723.9N

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

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At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:
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in the primitive yo-yo apparatus (figure 1), you replace the solid cylinder with a hollow cylinder of mass m , outer radius r ,
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The magnitude of the downward acceleration of the hollow cylinder is 6m/s^2.

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The rate at which an object's velocity with respect to time changes is called its acceleration. The direction of the net force imposed on an item determines its acceleration in relation to that force. According to Newton's Second Law, the magnitude of an object's acceleration is the result of two factors working together

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A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th
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Answer:

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