Question

What is the force needed to accelerate a 0.170 kg hockey puck from rest to 55.0 m/s over a period of 0.600 seconds?

Answer 1

Let's first find the acceleration of the hockey puck:

$$a=\dfrac{v-v_0}{t}=\dfrac{55.0\ \dfrac{m}{s}-0\ \dfrac{m}{s}}{0.6\ s}=92.0\ \dfrac{m}{s^2}.$$

Then, we can find the force needed to accelerate the 0.170 kg hockey puck from the Newton's Second Law of Motion:

$$F=ma=0.170\ kg\cdot 92.0\ \dfrac{m}{s^2}=15.64\ N.$$

Answer:

$F=15.64\ N.$