A mass m 1 = 3.4 kg rests on a frictionless table and connected by a massless st

Question

A mass m1= 3.4 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2= 4.6 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.89 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 = 4.6kg*9.8m/s^2*(0.89m) = 40.12J 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*40.12/(3.4+4.6)) = 3.16m/s 4.The work by tension on m1 = change in kinetic energy of m1 so W = 1/2*m1*v^2 = 1/2*3.4*3.16^2 = 17.05J 5. Since W = F*d then F (or tension) = W/d = 17.05/0.89 = 19.15N 6. Net work = answer from 1) = 40.12J

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