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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

370 N
370 N

80 N
80 N

70 N
70 N

380 N
380 N

The magnitude of the net force acting on the boy can be calculated by adding up all the forces acting on him.

The force exerted by the boy on the ice is 300 N.
The force exerted by his friend on the boy is 75 N.
The force of drag and friction is 5 N.

To calculate the net force, we add up these forces:

Net force = Boy's force + Friend's force + Drag and friction force
Net force = 300 N + 75 N + 5 N
Net force = 380 N

Therefore, the magnitude of the net force acting on the boy is 380 N.

Net horizontal force on boy = 75 - 5 = 70 Newtons

Force up on boy from ice = 300 N
gravity down on Boy = -300 N
Net vertical force on boy = 0 N

So I get 70 Newtons net force and it is horizontal

Apologies for the confusion. You are correct. The net horizontal force on the boy is indeed 70 Newtons. The force exerted by his friend (75 N) and the force of drag and friction (5 N) are in the same direction, resulting in a total net force of 70 Newtons acting horizontally on the boy. The forces acting vertically on the boy (force up from the ice and gravity pulling him down) cancel each other out, resulting in a net vertical force of 0 Newtons.

Thanks. I had a hunch the boy would not head for the moon or the bottom of the pond.

You're welcome! Yes, any net vertical force on the boy would result in him accelerating either upwards or downwards. In this case, the forces cancel each other out vertically, so the boy will not experience any vertical acceleration. His motion will be primarily in the horizontal direction, determined by the net horizontal force acting on him.