please do parts a-d Class Management| Help Homework #5 (Ch 5: Newton\'s laws of
ID: 1789839 • Letter: P
Question
please do parts a-d
Class Management| Help Homework #5 (Ch 5: Newton's laws of Motion) Begin Date: 12:00:00 AM 162 1712 00 0 AM Due Da 102720171 15 00 pMEndDatel (6%) Problem 15: Two blocks are connected by a massless rope The rope passes over an ideal (fnictionless and massless) pulley such that one block with mass m 14.25 kg is on a horizontal table and the other block with mass m-6.5 kg hangs vertically Both blocks experience gravity and the tension force, T Use the coordinate system specified in the diagram. im t Status re for view Status Completed Completesd Completed Partial Completedd theespertta.cam 25% Part (a) Assuming friction forces are negligible wite an expression, using only the variables provided for the a celeration that the block of mass my experiences in the x-direction Your answer should involve the tension. T -, a, 25% Part b) Under the same assumptions. for the accelerat direction Your answer should be it terms of the tension, T and m ns, wite an expression for the acceleration o, the block of mass w2 experiences in the y- Parrio Carcfully consdet how the accelerations a and a, are related Solve for the magnitude of the acceleration. a, of the block Partial Completed Completed of mass . in meters per square second. 25% Parti di Find the magnitude of the tension in the rope. in newtons Grade Seemar Potential Submissions L0144Explanation / Answer
a] For m1, the net force will be:
Fnet = T = m1a
=> a = a2 = T/m1
b]
For m2,
m2g - T = m2a
=> a2 = a = [m2g - T]/m2
c] T = m1a
therefore, a = [m2g - m1a]/m2
=> a = [m2/(m1+m2)]g
=> a = [6.5/(6.5+14.25)](9.8) = 3.07 m/s2
d] T = 14.25(3.07) = 43.746 N.
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