A mass m 1 = 3.3 kg rests on a frictionless table and connected by a massless st
ID: 1280112 • Letter: A
Question
A mass m1 = 3.3 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 4.4 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.87 m.
1) How much work is done by gravity on the two block system? J
2) How much work is done by the normal force on m1? J
3)What is the final speed of the two blocks? m/s
4)How much work is done by tension on m1? J
5)What is the tension in the string as the block falls? N
6)The work done by tension on only m2 is:
positive
zero
negative
7) What is the NET work done on m2?
A mass m1 = 3.3 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 4.4 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.87 m. 1) How much work is done by gravity on the two block system? J 2) How much work is done by the normal force on m1? J 3)What is the final speed of the two blocks? m/s 4)How much work is done by tension on m1? J 5)What is the tension in the string as the block falls? N 6)The work done by tension on only m2 is: positive zero negative 7) What is the NET work done on m2?Explanation / Answer
1.) Work done by gravity = m2g.d = 4.4x9.8x0.87 = 37.51 J
2.) work done by Normal force = 0 => because no displacement in the direction of normal force.
3.) To calculate final speeds we have to find out the accelertion of the system. let it be a m/s2 in the downward direction
& let tension in the String be T
So writing equations , T = m1a & m2g - T = m2a
=> m2g - m1a = m2a
=> a = 4.4x9.8/7.7 = 5.6 m/s2
So , final speed of m1 => v = sqrt(2as) = sqrt(2x5.6x0.87) = 3.12 m/s
final speed of m2=> 3.12 m/s
4) Work done by tension on m1 = m1xaxs = 16.07 J
5.) From above equations , Tension in the String = m1a = 18.48 N
6.) Work done by Tension on m2 is Negative , because displacement in opposite direction of Tension force.
7.) Net work done on m2 = netforceXdisplacement = m2aX0.87 = 21.43 J
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