A child whose weight is 297 N slides down a 6.60 m playground slide that makes an angle of 38.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.160. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.335 m/s, what is her speed at the bottom?

Given that,

Weight of child is W = 297N

Length of slide is 6.6m

Angle of inclination is 38°

Coefficients of kinetic friction is 0.16

A. Energy transferred to thermal heat is the workdone by friction

The frictional force is given as

Fr=μN

And the Normal (N) is given as

N=mgCos θ, since W=mg

N=WCos θ

N = 297Cos38

N = 234.04N

Then, the frictional force is

Fr=μN

Fr=0.16*234.04

Fr=37.45N

Now, the work done by friction is

W (friction) = frictional force * distance moved by body

Wf = Fr * d

Wf = 37.45 * 6.6

Wf = 247.15J

b. The change in K. E is given as

∆K. E = Wg - Wf

Wg Is the work done downward by the child

So the forward force on the child down the slope is given as

F=WSin θ

F=297Sin38

F=182.85N

Then, the work done forward is

Wg=F*d

Wg=182.85*6.6

Wg=1206.82J

K. E = Wg - Wf

½mv² = 1206.82 - 247.15

To get the mass of the child let use the weight formula

W=mg

m=W/g

m=297/9.81

m=30.28kg

½mv² = 1206.82 - 247.15

½*30.28v²=959.67

15.14v²=959.67

v²=959.67/15.14

v²=63.39

v=√63.39

v=7.96m/s