A mass m1 = 4.8 kg rests on a frictionless table and connected by a massless str

Question

A mass m1 = 4.8 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 4.5 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.78 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? 7) What is the NET work done on m2?

Explanation / Answer

1) How much work is done by gravity on the two block system?

answer) we have formula work done = m(2) * g* h = 4.5 kg * 9.81 m/s2 * 0.78 m = 34.4331 Joule =====answer)

2) How much work is done by the normal force on m1?

answer) Zero because we know that normal force is perpendicular to the motion.

3) What is the final speed of the two blocks?

answer) we can solve using work energy :

we have an equation :  v = sqrt(2*W/(m1 + m2))

put in the values we get

v = sqrt ( 2 * 34.43/ ( 4.8 + 4.5) = 2.72108 m/s =============answer)

4) How much work is done by tension on m1

answer)

work done = 0.5 * m1 * v2 = 0.5 * 4.8 kg * ( 2.72108)^ 2 = 17.7703 Joule =============answer)

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