Answer to Question #304144 in Physics for ameer

Question

A 72.5 kg

k

g

 skater moving initially at 2.40 m/s

m

/

s

 on rough horizontal ice comes to rest uniformly in 3.52 s

s

 due to friction from the ice.

A, What force does friction exert on the skater?

Express your answer with the appropriate units. Enter positive value if the direction of the force is the same as that of the initial velocity of the skater and negative value if the direction of the force is opposite.

Answer 1

Since the skater slows down, the force is opposite to the direction of the initial velocity. From the second Newton's law:

$$F = ma$$

where $m = 72.5kg$ and $a$ is the constant acceleration of the skater. The acceleration by definition is:

$$a = \dfrac{2.4m/s}{3.52s} \approx 0.682m/s^2$$

Thus, obtain:

$$F = 72.5kg\cdot 0.682m/s^2 \approx 49.4N$$

Answer. -49.4N.