A mass m1 = 5.7 kg rests on a frictionless table and connected by a massless str

Question

A mass m1 = 5.7 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 3.3 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.76 m. 1) How much work is done by gravity on the two block system? 2) How much work is done by the normal force on m1? 3) What is the final speed of the two blocks? 4) How much work is done by tension on m1? 5) What is the tension in the string as the block falls? 6) What is the NET work done on m2?

Explanation / Answer

1. The work done by gravity = m2*g*h = 3.3kg*9.8m/s^2*(0.76m) = 24.57J

2. Since the normal is perpendicular to the motion the work = 0

3. Use work-energy here so W = 1/2*(m1 + m2)*v^2

so v = sqrt(2*W/(m1 + m2)) = sqrt(2*24.57/(5.7+3.3)) = 2.33m/s

4.The work by tension on m1 = change in kinetic energy of m1

so W = 1/2*m1*v^2 = 1/2*5.7*2.33^2 =15.47J

5. Since W = F*d then F (or tension) = W/d = 15.47/0.76 = 20.35N

6. Net work = answer from 1) = 24.57J

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