Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

The 3.25kg collar shown below is attached to a spring and released from rest at

ID: 1844328 • Letter: T

Question

The 3.25kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.90m . The spring has a spring constant of k = 25.5N/m . The distance a is given as 1.10m . The datum for gravitational potential energy is set at the horizontal line through A and B.

Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at C is 3.59m/s and the spring has an unstretched length of 0.65m .

Express your answer to three significant figures and include the appropriate units.

The 3.25kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.90m . The spring has a spring constant of k = 25.5N/m . The distance a is given as 1.10m . The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at C is 3.59m/s and the spring has an unstretched length of 0.65m . Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

It is simply using energy conservation, change in energy = frictional energy

or

mgy - 1/2 mv2 -1/2 kx2 = frictional energy

frictional energy = 3.25*10*1.9 - 0.5*3.25*3.592 - 0.5*25.5*(1.54544984001001492- .452)

= 12.9364185981659966971 J

=> frictional force = energy / distance = 12.9364185981659966971/1.9

= 6.8086413674557877353158N

avg. frictional force = 6.809 N