Question

In: Physics

A child on a sled is travelling a total distance of 93 m down a dirt...

A child on a sled is travelling a total distance of 93 m down a dirt hill with an angle of 30 degrees from the horizontal from the top of the hill. The mass of the child and sled is 65 kg. The initial velocity is zero and the final velocity is 5.0 m/s. (a) What is the time it takes to travel down the hill? (b) What is the acceleration down the hill? (c) What is the force of friction? (d) What is the coefficient of kinetic friction?

Expert Solution

Gravitational acceleration = g = 9.81 m/s²

Angle of the slope = = 30^o

Length of the slope = L = 93 m

Height of the slope = H

H = LSin

H = (93)Sin30

H = 46.5 m

Mass of the child = m = 65 kg

Initial velocity of the child at the top of the hill = V₁ = 0 m/s

Final velocity of the child at the bottom of the hill = V₂ = 5 m/s

Friction force acting on the sled = f

The initial potential energy is converted into kinetic energy plus the work done against friction.

mgH = mV₂²/2 + fL

(65)(9.81)(46.5) = (65)(5)²/2 + f(93)

f = 310.09 N

Coefficient of kinetic friction =

Acceleration of the child down the hill = a

Time taken to travel down the hill = T

From the free body diagram,

mgCos = N

f = N

f = mgCos

310.09 = (65)(9.81)Cos(30)

= 0.561

ma = mgSin - f

(65)a = (65)(9.81)Sin(30) - 310.09

a = 0.134 m/s²

V₂ = V₁ + aT

5 = 0 + (0.134)T

T = 37.31 sec

a) Time taken to travel down the hill = 37.31 sec

b) Acceleration down the hill = 0.134 m/s²

c) Force of friction = 310.09 N

d) Coefficient of friction = 0.561

genius_generous answered 1 hour ago

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