Answer:
a) [tex]W_{fr} = 1129.602\,J[/tex], b) [tex]\mu_{k} = 0.268[/tex]
Explanation:
The seal is modelled after the Principle of Energy Conservation and the Work-Energy Theorem:
[tex]U_{g,A} + K_{A} = U_{g,B} + K_{B} + W_{fr}[/tex]
a) The work done by the kinetic friction is:
[tex]W_{fr} = U_{g,A}-U_{g,B}+K_{A}-K_{B}[/tex]
[tex]W_{fr} = (116\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.85\,m-0\,m) + \frac{1}{2}\cdot (116\,kg)\cdot \left[(0\,\frac{m}{s} )^{2} - (4.10\,\frac{m}{s} )^{2}\right][/tex]
[tex]W_{fr} = 1129.602\,J[/tex]
b) The coefficient of kinetic friction is:
[tex]W_{fr} = \frac{\mu_{k}\cdot (m\cdot g \cdot \cos \theta)\cdot (h_{A}-h_{B})}{\sin \theta}[/tex]
[tex]\mu_{k} = \frac{W_{fr}\cdot \sin \theta}{(m\cdot g \cdot \cos \theta)\cdot (h_{A}-h_{B})}[/tex]
[tex]\mu_{k} = \frac{1129.602\,J\cdot \sin 26.5^{\textdegree}}{(116\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (\cos 26.5^{\textdegree})\cdot (1.85\,m)}[/tex]
[tex]\mu_{k} = 0.268[/tex]
Answer:
(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
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.
[tex]U_i + K_i = U_f + K_f + W_{friction}[/tex]
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;
[tex]U_i = k_f + W_{friction}[/tex]
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
