As shown in the figure above, blocks A and B are connected by a massless string
ID: 2200661 • Letter: A
Question
As shown in the figure above, blocks A and B are connected by a massless string that passes over the outer edge of a pulley that is a uniform solid disk. Block A has a mass of 4.00 kg. Block B has a mass of 6.00 kg. The pulley has a mass of 3.00 kg. When the system is released from rest, it experiences a constant (and non-zero) acceleration. There is no friction between block A and the surface. Use g = 10.0 m/s^2.
What is the tension in the part of the string that is connected to block A?
Explanation / Answer
you must incorporate the rotational kinetic energy of the pulley! Write an equation for each object, add the three equations together, and solve for v. initially, the entire system is at rest, and we can ignore the gravitational potential energy of the block on the table (block A, I’ll call it m1), its equation is: 0 = 0.5m1v²----------------------->(1) The hanging block has some initial GPE and zero initial KE, but after falling 2.00 meters, we’ll say that all its GPE (relative to where it starts from) has been converted entirely to KE. Its equation is: m2gh = 0.5m2v²------------------>(2) The pulley’s initial RKE is zero, but after the hanging block has dropped the 2.00 meters, it has some RKE expressed as one-half the product of its moment of inertia and its angular velocity: 0 = 0.5I?² But I for a solid cylinder is 0.5MR² and ? may be written as v/R, so the equation becomes: 0 = 0.25Mv²---------------------->(3) Added together and solved for v, we get: V = v[2m2gh / (m1 + m2 + 0.5M)] = [2(3.00kg)(10.0m/s²)(2.00m) / (8.00kg + 3.00kg + 2.00kg)] = 3.038m/s Hope this helps. EDIT: Well, I see where I failed, and you should have too. Half of 2.00kg is 1.00kg, that's the correct last term in the denominator. The answer should then be 3.16m/s. It checks with the 2nd law.
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