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? (a) 70 N (b) 370 N (c) 80 N (d) 380 N

Answer 1

The answer is a hope it helps

Answer 2

Choice a.
`70\; \rm N`, assuming that the skating rink is level.

Step-by-step explanation:

Net force in the horizontal direction

There are two horizontal forces acting on the boy:

• The pull of his friend, and
• Frictions.

The boy should be moving in the direction of the pull of his friend. The frictions on this boy should oppose that motion. Therefore, the frictions on the boy would be in the opposite direction of the pull of his friend.

The net force in the horizontal direction should then be the difference between the pull of the friend, and the friction on this boy.

`\text{Net force, horizontal} = 75\; \rm N - 5\; \rm N = 70\; \rm N`.

Net force in the vertical direction

The net force on this boy should be zero in the vertical direction. Consider Newton's Second Law of motion. The net force on an object is proportional to its acceleration. In this question, the net force on this boy in the vertical direction should be proportional to the vertical acceleration of this boy.

However, because (by assumption) the ice rink is level, the boy has no motion in the vertical direction. His vertical acceleration will be zero. As a result, the net force on him should also be zero in the vertical direction.

Net force

Therefore, the (combined) net force on this boy would be:

`√((70\; \rm N)^2 + (0\; \rm N)^2) = 70\; \rm N`.