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## Homework Statement

A child is tobogganing down a hillside. The child and the toboggan together have a mass of 50.0kg. The slope is at an angle of 30.0° to the horizontal.

Assume that the positive y-direction is pointing in the direction of the normal force. Assume that the positive x-direction is down the incline.

Find the acceleration of the child

a) in the case where there is no friction

b) if the coefficient of friction is 0.15

## The Attempt at a Solution

a) Fn + FgI = mgcosθ

= (50)(9.8)cos 30°

= 424N

Fnet = FgII = mgsinθ

= (50)(9.8)sin 30°

= 245 N (+ x-direction)

Fnet = ma

a = Fnet/m

= 245/50

= +4.9m/s/s

Therefor the child's acceleration in the case where there is no friction is 4.9 m/s/s.

b) FgI = 424 (- y-direction)

FgII = 245 (+ x-direction)

Fnet = FgII (+ x-direction) + Fk (- x-direction)

Fn = FgI = 424N (down)

Fk = μkFn

= (0.15)(424)

= 63.6N (- x-direction)

Fnet = FgII + Fk

= (+245)+(-63.6)

= +181.4N

Fnet =ma

a = Fnet/m

= 181.4/50

= +3.63m/s/s

Therefore the child's acceleration is +3.63 m/s/s if the coefficient of friction is 0.15.