A42.0 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?

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

the answer is the index of refraction of the second medium is higher. i just took the test.