Question

A 116 kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.85 m higher than the surface of the water and the ramp is inclined at an angle of 26.5 ∘ above the horizontal.

Part A If the seal reaches the water with a speed of 4.10 m/s, what is the work done by kinetic friction?

Part B What is the coefficient of kinetic friction between the seal and the ramp?

Answer 1

(A) Work done by friction, Wf = 1128.1 J

(B) The coefficient of kinetic friction between the seal and the ramp,μ is 0.268

Step-by-step explanation:

Given;

mass of seal, m = 116 kg

top of the ramp, h = 1.85 m

angle of inclination, θ = 26.5⁰

Part A

Apply the principle of conservation of Energy;

sum of initial potential energy and initial kinetic energy = sum of final potential energy, final kinetic energy and work lost due to friction.

`U_i + K_i = U_f + K_f + W_(friction)`

Initial kinetic energy is zero, since the seal slides from rest.

Final potential energy is zero, at the end of the ramp, height is zero.

The equation above reduces to;

`U_i = k_f + W_(friction)`

Mgh = ¹/₂Mv² + Wf

Wf = Mgh - ¹/₂Mv²

Wf = (116 x 9.8 x 1.85) - (¹/₂ x 116 x 4.1²)

Wf = 2103.08 - 974.98 = 1128.1 J

Thus, work done by friction, Wf = 1128.1 J

Part B

work done by friction, Wf = Frictional force x distance moved by the seal down the slope

Wf = Fk x d

Wf = μmgcosθ x d

d is the slope of the inclined ramp to the horizontal, this calculated using trigonometry ratio;

d (slope) = h/sinθ

d = 1.85/sin26.5

d = 4.146 m

Wf = μmgcosθ x d

1128.1 = μ x 116 x 9.8 x cos(26.5) x 4.146

1128.1 = μ x 4217.98

μ = 1128.1 / 4217.98

μ = 0.268

Thus, the coefficient of kinetic friction between the seal and the ramp is 0.268