A 23 kg child goes down a straight slide inclined 38 above horizontal. The child

Question

A 23 kg child goes down a straight slide inclined 38 above horizontal. The child is acted on by his weight, the normal force from the slide, kinetic friction, and a horizontal rope exerting a 30 degree force. How large is the normal force of the slide on the child? Express your answer using two significant figures.

Explanation / Answer

always eager to do a problem when the DIAGRAM is supplied ! :>) Normal force = component of child's weight that acts normal to slide - 30 sin 38° Normal force = mg(cos 38°) - 30(sin 38°) Normal force = (23)(9.8)(cos 38) - 30(sin 38) = 178 - 18.5 = 159.5 N ANS

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