Question

A 92.71 kg skater moving initially to the right at 0.39 m/s on rough horizontal ice comes to rest uniformly (constant acceleration) in 5.85 s due to friction from the ice. What force does friction exert on the skater?

Answer 1

Sol::

Given::

Mass(m)= 92.71 kg
Speed(Vxi)= 0.39 m/s
Time(t)= 5.85 s

By figuring out the acceleration assuming it is uniform:
We have

a = (Vxf-Vxi)/t
= (0-0.39)/5.85
= -0.0666 m/s^2

Now using Newton's second law

Ff = ma
= 92.71*(-0.0666)
= -6.174 N

Here the negative sign shows the frictional force oppositeto the movement of skater

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Answer 2

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