A mass m1 = 3.9 kg rests on a frictionless table and connected by a massless str

Question

A mass m1 = 3.9 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 5.7 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.8 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) What is the NET work done on m2?

Explanation / Answer

i have consider 3.9=6, so be sure. work done by gravity on m2 is SHARED among both m1 and m2. Sure gravity may've done positive work on m2 as m2 falls, but also the TENSION did negative work on m2, as it restrained m2 from falling as fast as gravity alone would allow. To solve this problem, construct a kinetic energy equation for both bodies. Set the speeds equal, because they are constrained by the inextensible string. KE1 = 1/2*m1*v1^2 KE2 = 1/2*m2*v2^2 v1 = v2 Then study the source of the energy (the loss of GPE), and SHARE it among both KE1 and KE2. dGPE2 = KE1 + KE2 m2*g*d = 1/2*m1*v1^2 + 1/2*m2*v2^2 v1 = v2 Replace v1 and v2 with v: m2*g*d = 1/2*m1*v^2 + 1/2*m2*v^2 Factor: m2*g*d = 1/2*(m1 + m2)*v^2 Solve for v: v = sqrt(2*m2*g*d/(m1 + m2)) Now find the kinetic energy of body 2: KE2 = 1/2*m2*v^2 KE2 = 1/2*m2*(2*m2*g*d/(m1 + m2)) Simplify: KE2 = m2^2 * g*d/(m1 + m2) Now make an argument why this will be your final answer. The net work done accelerating an object from rest equals the kinetic energy it possesses once in motion. Thus, the answer to Wnet2: Wnet2 = KE2 Wnet2 = m2^2 * g*d/(m1 + m2) Data: m1 = 6 kg m2 = 3.3 kg g = 9.8 N/kg d =0.87 m Result: Wnet = +9.98 Joules Why positive? Because the net work done ON the object caused it to INCREASE in speed.

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