Question

If a 90 kg hockey player slows from 18 m/s to 12 m/s over 0.6 s, what is the force exerted by his skates on the ice? A. -540 N B. -900 kg-m/s2 C. -1,080 kg-m/s D. -1,800 N

Answer 1

Answer:

B. -900 kg-m/s2

Explanation:

First of all, we need to calculate the acceleration of the hockey player, which is given by:

$a=\frac{v-u}{t}$

where

v = 12 m/s is the final speed

u = 18 m/s is the initial speed

t = 0.6 s is the time

Substituting into the formula, we find

$a=\frac{12 m/s-18 m/s}{0.6 s}=-10 m/s^2$

And now we can calculate the force exerted by using Newton's second law:

$F=ma$

where m=90 kg is the mass of the hockey player. Substituting into the formula, we find

$F=(90 kg)(-10 m/s^2)=-900 N$