Question

A skater starts from rest v1=0, from the top of an h=6m hill and travels d=20m until it gets to the bottom of the hill where its speed is v2=1.2m/s. If the skater has m=50kg, what is the friction force?

Answer 1

The friction force is 1.5N

• skater's initial velocity v1=0

• height of the hill=20m

• skater's final velocity=1.2m/s

• Mass of the skater=50kg

Now, using the third equation of motion-

$v^{2}= u^{2} +2as$

v=final velocity

u=initial velocity

a=acceleration

s=distance

substitute the values in the above equation

1.2=0+ 2a×20

a=1.2/40

a=0.03

friction force=mass ×acceleration

                    =50×0.03

                    =1.5N

hence, The friction force is 1.5N

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