Question

a boy who exerts a 300 - n force on the ice of a skating rink is pulled by his friend with a force of 75 n, causing the boy to accelerate across the ice. if drag and the friction from the ice apply a force of 5 n on the boy, what is the magnitude of the net force acting on him? (1 point)
80 n
70 n
380 n
370 n

Explanation:

Step1: Determine the direction of forces

The pulling force is in the direction of motion ($F_{pull}=75\ N$), and the drag - friction force is opposite to the direction of motion ($F_{drag - friction}=5\ N$). The normal force (the boy's force on ice, $300\ N$) and the weight of the boy (not relevant for net force in the direction of motion as they cancel each other out in the vertical direction, assuming no vertical acceleration) are not considered for the net force in the horizontal (direction of motion) direction.

Step2: Calculate the net force in the direction of motion

Using the formula for net force $F_{net}=F_{pull}-F_{drag - friction}$.
Substitute the values: $F_{net}=75 - 5$.

Answer:

$70\ N$