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)
○ 80 n
○ 70 n
○ 380 n
○ 370 n

Explanation:

Step1: Identify forces in the same direction

The boy is pulled with a force of 75 N, and we assume this is in the direction of motion. The drag and friction force (5 N) is opposite to the direction of motion. Wait, actually, the 300 - N force is the force the boy exerts on the ice, but by Newton's third law, the ice exerts a 300 - N force on the boy in the direction of motion? Wait, no, the problem says the boy is pulled by his friend with 75 N, and drag/friction is 5 N. Wait, maybe I misread. Wait, the boy exerts 300 N on the ice, so the ice exerts 300 N on the boy (action - reaction). But then he is pulled by friend with 75 N? Wait, no, maybe the 300 N is the force from the ice (thrust) and the friend pulls with 75 N, and friction is 5 N. Wait, no, let's re - read: "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?"

Wait, when the boy exerts a 300 - N force on the ice, the ice exerts a 300 - N force on the boy (Newton's third law) in the direction of motion. Then his friend pulls him with 75 N (also in the direction of motion), and drag/friction is 5 N opposite to the direction of motion.

So forces in the direction of motion: \(F_{1}=300\ N\) (from ice) and \(F_{2} = 75\ N\) (from friend). Force opposite: \(F_{f}=5\ N\) (friction/drag).

Step2: Calculate net force

Net force \(F_{net}=(F_{1}+F_{2})-F_{f}\)

Substitute the values: \(F_{1} = 300\ N\), \(F_{2}=75\ N\), \(F_{f}=5\ N\)

\(F_{net}=(300 + 75)-5\)

First, calculate \(300+75=375\ N\)

Then, \(375 - 5=370\ N\)? Wait, no, wait, maybe I got the forces wrong. Wait, maybe the 300 - N force is not a force on the boy. Wait, the boy exerts 300 N on the ice, so the ice pushes him with 300 N (forward). The friend pulls him with 75 N (forward), and friction is 5 N (backward). So total forward force: \(300 + 75=375\ N\), backward force: 5 N. Net force: \(375-5 = 370\ N\)? But the options have 370 N as an option. Wait, but maybe I misinterpreted the 300 - N force. Wait, maybe the boy is skating, and the 300 - N force is not a force on him? Wait, no, Newton's third law: when he pushes the ice with 300 N, the ice pushes him with 300 N. Then friend pulls with 75 N, friction is 5 N. So net force is \(300 + 75-5=370\ N\).

Wait, but let's check again. Maybe the 300 - N force is the force he applies, but the force on him from the ice is 300 N. Then friend's force is 75 N (same direction), friction is 5 N (opposite). So net force is \(300 + 75-5=370\ N\).

Answer:

370 N (the option: 370 N)