A slide inclined at 35 takes bathers into a swimming pool. with water sprayed on the slide the bathers spends only one third as much timeo n the slide as whne its dry. find the coefficient of friction on the dry slide

Refer to the figure shown below.

mg = the weight of the person on the slide.

N = mg cos (35°) = 0.8192 mg, the normal reaction

F = mg sin (35°) = 0.5736 mg, the force acting down the slide.

R = μN = 0.8192 μmg, the resisting force due to friction

μ = the coefficient of dynamic friction on the dry slide.

d = the length of the slide.

g = 9.8 m/s²

All measurements are in SI units.

We shall consider two cases.

Case 1 (The slide is lubricated by water spray).

Assume that the coefficient of friction is approximately zero.

Let a = the acceleration down the slide. Then

0.5736 mg = ma

a = 0.5736g = 5.6213 m/s²

Let t = the time to travel the length of the slide from rest.

d = (1/2) * (5.6213 m/s²) (t s) ² = 2.8106t² m (1)

t² = 0.3558d

t = 0.5965√d s (2)

Case 2 (The slide is dry)

The time to travel down the slide is 3 * (0.5965√t) = 1.7895√t s.

Let a = the acceleration. Then

0.5796mg - 0.8192μmg = ma

a = 0.5796g - 0.8192μg = 5.6801 - 8.0282μ m/s²

The travel doen the slide from rest is given by

d = (1/2) * (5.6801 - 8.0282μ) * (1.7895√t) ²

d = (9.0947 - 12.8544μ) t² (3)

Equate (1) and (3).

9.0947 - 12.8544μ = 2.8106

12.8544μ = 6.2841

μ = 0.489

Answer: μ = 0.49 (nearest hundredth)