Answer to Question #157075 in Mechanics | Relativity for Madison

Question

A 500 N boy is standing on frictionless roller skates. If someone pulls him, horizontally, with 40 N of force to the North, what will be his acceleration?

Answer 1

Let's first find the mass of the boy:

$$m=\dfrac{W}{g}=\dfrac{500\ N}{9.8\ \dfrac{m}{s^2}}=51\ kg.$$

Then, we can find the acceleration of the boy from the Newton's Second Law of Motion:

$$F=ma,$$ $$a=\dfrac{F}{m}=\dfrac{40\ N}{51\ kg}=0.78\ \dfrac{m}{s^2}.$$

Answer:

$a=0.78\ \dfrac{m}{s^2}.$