The 3.15 kg collar shown below is attached to a spring and released from rest at
ID: 1366906 • Letter: T
Question
The 3.15 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 2.10 m . The spring has a spring constant of k = 21.5 N/m . The distance a is given as 1.25 m . The datum for gravitational potential energy is set at the horizontal line through A and B.
1. Determine the magnitude of the velocity of the collar at C when the effects of friction are neglected. The spring is initially unstretched when the collar is at A.
2. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at C is 3.36 m/s . The spring is initially unstretched when the collar is at A.
3. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at C is 3.36 m/s and the spring has an unstretched length of 0.65 m .
Explanation / Answer
extension of the spring when the collar is at point C, x = sqrt(a^2 + y^2) - a
= sqrt(1.25^2 + 2.1^2) - 1.25
= 1.194 m
1) Apply conservation of energy
initial mechanical energy = final mechanical energy
m*g*y = 0.5*k*x^2 + 0.5*m*v^2
0.5*m*v^2 = m*g*y - 0.5*k*x^2
v = sqrt(2*g*y - k*x^2/m)
= sqrt(2*9.8*2.1 - 21.5*1.194^2/3.15)
= 5.6 m/s
2) Workdone by friction change in mechinaical energy
F*d = 0.5*m*(v^2 - v'^2)
F = 0.5*m*(v^2 - v'^2)/d
= 0.5*3.15*(5.6^2 - 3.36^2)
= 31.6 N
3)
initial extension of the spring, x1 = 1.25 - 0.65 = 0.6 m
final extension, x2 = 1.194 m
Again Apply conservation of energy
initial mechanical energy = final mechanical energy
m*g*y = 0.5*k*(x2^2 x1^2) + 0.5*m*v^2
0.5*m*v^2 = m*g*y - 0.5*k*(x2^2 - x1^2)
v = sqrt(2*g*y - k*(x2 - x1^2)^2/m)
= sqrt(2*9.8*2.1 - 21.5*(1.194^2 - 0.6^2)/3.15)
= 5.82 m/s
Workdone by friction change in mechinaical energy
F*d = 0.5*m*(v^2 - v'^2)
F = 0.5*m*(v^2 - v'^2)/d
= 0.5*3.15*(5.82^2 - 3.36^2)
= 35.57 N
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