- #1

StephenPrivitera

- 363

- 0

Find the work done by gravity when two masses (42kg and 75kg) which hang vertically downward are connected by a pulley. The masses are initially at rest. The mass move under the influence of gravity for 2.0s. Assume no friction in the pulley.

My solution:

Considering the larger mass m1, the work done by gravity on m1 is

W1=w1*d, where W1 is the work done on m1 by gravity, w1 is the force of gravity acting on m1, and d is the distance traversed by m1 in the time period. d is the same for both m1 and m2 (the smaller mass) since both move under the influence of gravity and therefore have the same acceleration. The net force on m1 is the force of gravity on m1 minus the tension of the rope (if downward is to be the positive direction). The tension of the rope is equivalent in magnitude to the weight of m2. Thus, F(net)1=g*(m1-m2)=m1*a. This distance traversed by m1 is then d=g/(2m1)*(m1-m2)(t^2). The work done by gravity on m1 is then

W1=g*m1*g/(2m1)*(m1-m2)(t^2)=(g^2)/2*(m1-m2)(t^2)=6300J

Without calculating, I can see that the work done by gravity on the other mass should be equivalent in magnitude to this answer.

Book's answer: 1.8x10^3J

Where did I go wrong?

Thanks in advance for the help.

My solution:

Considering the larger mass m1, the work done by gravity on m1 is

W1=w1*d, where W1 is the work done on m1 by gravity, w1 is the force of gravity acting on m1, and d is the distance traversed by m1 in the time period. d is the same for both m1 and m2 (the smaller mass) since both move under the influence of gravity and therefore have the same acceleration. The net force on m1 is the force of gravity on m1 minus the tension of the rope (if downward is to be the positive direction). The tension of the rope is equivalent in magnitude to the weight of m2. Thus, F(net)1=g*(m1-m2)=m1*a. This distance traversed by m1 is then d=g/(2m1)*(m1-m2)(t^2). The work done by gravity on m1 is then

W1=g*m1*g/(2m1)*(m1-m2)(t^2)=(g^2)/2*(m1-m2)(t^2)=6300J

Without calculating, I can see that the work done by gravity on the other mass should be equivalent in magnitude to this answer.

Book's answer: 1.8x10^3J

Where did I go wrong?

Thanks in advance for the help.

Last edited by a moderator: