Question

In: Physics

A mass m1 = 4.7 kg rests on a frictionless table and connected by a massless...

A mass m1 = 4.7 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 4.3 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 on the string as the block falls?

6. What is the NET work done on m2?

Solutions

Expert Solution

1. The only mass the gravity do work is m2 over (m1 is allways on the table) then, the work done by gravity is

     (1)

(positive because the mass falls.)

2. The normal force is, by definition, allways perpendicular to the table and to the movement of m1, then is work is zero.

3. Both masses move with the same speed. The only external forces that produces work on the system is the gravity (calculated in 1.). Using the Work-Energy theorem:

      (2)

Where Wext = Wg and Ec is the kinetic energy of all the system. Eci = 0, because the system is initially at rest. So:

        (3)

solving for vf:

       (4)

4. The work done by tension on m1 is

          (5)

To calculate the work of the tension, first we have to calculate the tension.

5. To know the tension we have to use the 2nd Newton law in each mass. In m1:

,                (6)

in m2:

               (7)

Substituting T from (6) in (7) and solving for a we have:

.           (8)

Substituting this result in (6):

,        (9)

Now we are able to complete the answer in point 4. Substituting (9) in eq. (5):

6. The net work on m2 is done by both forces acting on it:

(The work of T is negative because T points upwards and the displacement is downwards.)


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