A 42.0 kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.95 m higher than the surface of the water and the ramp is inclined at an angle of 35.0 degrees above the horizontal. A) If the seal reaches the water with a speed of 4.30 m/s, what is the work done by kinetic friction? B) What is the coefficient of kinetic friction between the seal and the ramp?

Question

Brooklyn

Physics

19 September, 10:40

A 42.0 kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.95 m higher than the surface of the water and the ramp is inclined at an angle of 35.0 degrees above the horizontal. A) If the seal reaches the water with a speed of 4.30 m/s, what is the work done by kinetic friction? B) What is the coefficient of kinetic friction between the seal and the ramp?

+1

Answers (1)

Know the Answer?

Paisley Rios · Answer 1

A) Wf = - 414.33 J

B) μk=0.36

Explanation:

Newton's second law

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=42.0 kg mass of the seal

h = 1.95 m : hight of the ramp

θ = 35° angle of the ramp with respect to the horizontal

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the seal

We define the x-axis in the direction parallel to the movement of the seal on the ramp and the y-axis in the direction perpendicular to it.

W: Weight of the seal : In vertical direction downward

FN : Normal force : perpendicular to the direction the ramp

fk : Friction force: parallel to the direction to the ramp

Calculated of the weight of the seal

W = m*g = (42 kg) * (9.8 m/s²) = 411.6 N

x-y weight components

Wx = Wsin θ = (411.6) * sin (35) °=236.08 N

Wy = Wcos θ = (411.6) * cos (35) ° = 337.16 N

Calculated of the Normal force

∑Fy = m*ay ay = 0

FN-Wy = 0

FN=Wy = 337.16 N

Calculated of the Friction force:

fk=μk*FN = μk * 337.16 N Equation (1)

A) Principle of work and energy

ΔE = Wf

ΔE:mechanical energy change

Wf: Work done by kinetic friction force

K : Kinetic energy

U: Potential energy

ΔE = Wf

Ef-Ei = Wf

(K+U) final - (K+U) initial = Wf

((1/2) mv²+0) - (0+m*g*h) = Wf

(1/2) (42) (4.3) ² - (42) (9.8) (1.95) = Wf

388.29-802.62 = Wf

Wf = - 414.33 J

B) The coefficient of kinetic friction between the seal and the ramp

Wf = f*d Equation (2)

d: length of the ramp

sin θ = h/d

d = h / sin θ = 1.95 / sin 35

d = 3.4 m

We replace Wf = - 414.33 J and d = 3.4 m in the equation (2)

-414.33 J = - f * (3.4m)

fk = 414.33 N*m / 3.4m

fk=121.86 N

Calculated of the coefficient kinetic friction

μk=fK/FN

μk=121.86 N/337.16 N

μk=0.36