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Secure I https//session. mastering physics.com/myct/itemview?assignmentProblemIDE75108687 University Physics at FDU Signed in as brianna thomas Help Close UPS17-MP1 Problem 12.69 Resources e previous l 4 of 7 l next The two blocks in the figure(Eigure 1) are connected by Part A a massless rope that passes over a pulley. The pulley da is 17 cm in diameter and has a mass of 2.5 kg As the If the blocks are released from rest, how long does it take the 4.0 kg pulley turns, friction at the axle exerts a torque of block to reach the floor magnitude 0.53 N.m Express your answer to two significant figures and include the appropriate units. Figure 1 of 1 Value Units Submit My Answers Give Up 4.0 kg Continue 1,0 m 2.0 kg

Explanation / Answer

Consider the three objects (block 1, block 2, and the pulley) separately. Also recall that because the rope doesn't slip, the acceleration of each object is equal, we just have to be careful about the signs.

Write sum of torques about axis of pulley (f is the torque of the axle friction):
R*T1 - R*T2 - f = I*alpha

Use sum of forces for block 1 (4.0 kg) to get expression for T1, being careful to consider sign of acceleration (a is down):
T1 - m1g = m1(-a) => T1 = m1g - m1a

Use sum of forces for block 2 (2.0 kg) to get expression for T2, being careful to consider sign of acceleration (a is up):
T2 - m2g = m2a => T2 = m2g + m2a

Substitute for T1 and T2 into sum of torques and recall alpha = a/R, and I for a disk = 0.5MR^2, being careful to consider sign of pulley's acceleration (counterclockwise is positive)
R(m1g - m1a) - R(m2g + m2a) - f = (0.5MR^2)(a/R)

Simplify and solve for a:
a = (Rm1g - Rm2g - f) / (0.5MR + Rm1 + Rm2) = 1.136/0.61625 = 1.843

Solve for time from simplified linear kinematics equation:
t = (2h/a)^0.5

a = 1.843 m/s^2
t = 1.04 s

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