Question

A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. at the bottom, she is moving 20.3 m/s.how much work did friction do on the woman?(unit=J)PLEASE HELP

Answer 1

Answer:

W = 21409.2 J

Explanation:

Given,

The mass of the woman, m = 77 Kg

The length of the water slide, S = 42.6 m

The inclination of the water slide, ∅ = 42.3

The constant velocity of the women sliding, 20.3 m/s

The kinetic friction of the sliding force is given by the formula

                               Fₓ = μₓ η

Where,

                    μₓ - coefficient of kinetic friction

                     η - normal force acting on the body

Since the water slide is inclined at an angle and the person is sliding with constant velocity. The coefficient of friction becomes,

                     μₓ = tan∅

And,                η = mg cos∅

Therefore, the kinetic friction force becomes

                          Fₓ =  tan∅  mg cos∅

Substituting the given values in the above equation

                           Fₓ = 0.9 x 77 x 9.8 x 0,74

                               = 502.56 N

The work done by the kinetic friction on the person

                            W = Fₓ · S    

                                = 502.56 N x 42.6 m

                                = 21409.2 J

Hence, the work done by the friction on the woman is, W = 21409.2 J