A child whose weight is 228 N slides down a 4.10 m playground slide that makes an angle of 44.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.0720. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.307 m/s, what is her speed at the bottom?

a) 48.42J

b) The velocity at the bottom of the slide will be 0.307m/s

Explanation:

There is no conservation of energy when friction is invoved. Since the child is accelerating at a constant velocity and the force on the slide is in the same direction as the direction of the motion. The net force of the child is balanced

FN = W cos theta

FN = 228 × cos44= 164.0N

◇ET = Uk × FN× d

Where UK = coefficient of static friction = 0.0720

d = distance

Substituting these values into the equation

◇ET = 0.0720 × 164.0 × 4.10

◇Et = 48.42J

b) Using kinetic Energy equation

1/2 × MV1^2 = 1/2× MV2^2

But W = mg

M = W/g = 228 /9.81 = 23.27kg

1/2 × 23.27^2 = 1/2 × 23.27 ×V2^2

11.64 × 0.094 = 11.64V2^2

1.097 = 11.64V2^2

V2^2 = 1.097/11.64

V = Sqrt(0.094)

V = 0.3069 approximately 0.307m/s