Question

net force and acceleration quick check
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)
380 n
70 n
370 n
80 n

Explanation:

Step1: Determine the direction of forces

The force exerted by the friend is in the direction of motion (let's take it as positive), and the drag - friction force is opposite to the direction of motion (negative). The 300 - N force the boy exerts on the ice is the normal force (perpendicular to the direction of motion, so it doesn't contribute to the net force in the direction of motion).

Step2: Calculate the net force

Using the formula for net force \(F_{net}=F_{pull}-F_{drag - friction}\), where \(F_{pull} = 75\space N\) and \(F_{drag - friction}=5\space N\).

$$F_{net}=75 - 5$$

$$F_{net}=70\space N$$

Answer:

70 N