Q.445. Work/energy.

Question 445.
An 18 kg girl slides down a playground slide that is 3.6 m high.
When she reaches the bottom of the slide, her speed is 1.3 m/s.
a)How much energy was dissipated by friction?
b)If the slide is inclined at 20°, what is the coefficient of friction between the girl and the slide?

Answer 445.
a)

Energy dissipated by friction
= her initial PE - her final KE
= mgh - (1/2)mv^2
= 18 * [9.81 * 3.6 - 0.50 * (1.3)^2]
= 18 * (35.316 - 0.845) J
= 629.48 J

b)
Normal force = 18 * 9.81 * cos20°
=> frictional force = μ * 18 * 9.81 * cos20°
=> work done by the frictional force
= (3.6 / sin20°) * μ * 18 * 9.81 * cos20°
=> (3.6 / sin20°) * μ * 18 * 9.81 * cos20° = 629.48
=> μ = 629.48 * tan20° / (3.6 * 18 * 9.81)
=> μ = 0.36.

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