A mass m1 = 5.2 kg rests on a frictionless table and connected by a massless str

Question

A mass m1 = 5.2 kg rests on a frictionless table and connected by a massless string to another mass m2 = 4.2 kg. A force of magnitude F = 34.0 N pulls m1 to the left a distance d = 0.87 m.
1) How much work is done by the force F on the two block system?
J
2) How much work is done by the normal force on m1 and m2?
J
3) What is the final speed of the two blocks?
m/s
4) How much work is done by the tension (in-between the blocks) on block m2?
J
5) What is the tension in the string?
N
6) What is the NET work done on m1?

Explanation / Answer

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

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