Question

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)
Responses

70 N
70 N

80 N
80 N

380 N
380 N

370 N

Answer 1

To find the magnitude of the net force acting on the boy, we can use Newton's second law of motion, which states that the net force is the sum of all forces acting on an object.

In this case, we have the following forces:

1. The boy exerts a force of 300 N.
2. His friend pulls him with a force of 75 N in the same direction.
3. The drag and friction force opposing the motion is 5 N.

We can sum these forces:

The total force applied by the boy and his friend:
\[
F_{\text{applied}} = 300 \, \text{N} + 75 \, \text{N} = 375 \, \text{N}
\]

Now, we subtract the opposing force (drag and friction) from the total applied force:
\[
F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}
\]
\[
F_{\text{net}} = 375 \, \text{N} - 5 \, \text{N} = 370 \, \text{N}
\]

Thus, the magnitude of the net force acting on him is **370 N**.

The correct response is:
**370 N**.