Answer:
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
