Answer:
Explanation:
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
