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)

Question

Responses

80 N
80 N

70 N
70 N

370 N
370 N

380 N

Answer 1

To find the net force acting on the boy, we need to consider the forces acting upon him. The forces involved are:

1. The boy's own force exerted on the ice: 300 N (in one direction, let's say positive)
2. The force exerted by his friend pulling him: 75 N (in the same positive direction)
3. The drag and friction force acting against him: 5 N (in the negative direction)

We can calculate the net force using the following formula:

\[ \text{Net Force} = (\text{Boy's Force}) + (\text{Friend's Force}) - (\text{Friction/Drag Force}) \]

Substituting in the values:

\[ \text{Net Force} = 300, \text{N} + 75, \text{N} - 5, \text{N} \] \[ \text{Net Force} = 375, \text{N} \]

So, the magnitude of the net force acting on him is 375 N. However, this response is not listed in your options. The nearest response provided was 370 N, but that value is incorrect based on the calculations. The correct calculation gives a net force of 375 N. If we strictly adhere to your options, the closest response would be best treated as 370 N, but please note that the accurate calculated value is 375 N.

If you need to respond from the provided options based solely on the correct answer being missing, the best choice would be 370 N.