Set 3 Problem number 13

A mass of 9 kg on a level tabletop is attached by a light string to a mass of 6 kg; the second mass is hanging vertically from a pulley. The string runs from the first mass in the horizontal direction to the pulley. The pulley is considered to be light and frictionless.

If there is no friction anywhere in the system:

What is the net force on the system?

As the system moves 2.1 meters, by how much does its gravitational potential energy change, and by how much does its kinetic energy change?
If the system is moving in the direction of the net force at 3.4 m/s, by how much will its velocity increase as it moves 2.1 meters?

Answer the same questions if the only friction in the system is on the mass on the tabletop, and if the coefficient of friction between the mass and the tabletop is .2.

Solution

We will throughout the solution consider the positive direction to be that in which the system accelerates.

In the absence of friction:

The gravitational potential energy of the system will decrease as the altitude of the hanging mass decreases, by an amount equal to the work done by gravity on the mass (note that the negative change in PE is equal to the work done by the system against gravity, since the work done by the system against an applied force is equal and opposite to the work done on the system by that force).

Since the force of gravity on the hanging mass is 58.8 Newtons, and since this force is in the direction of motion, we see that the work done by gravity on the mass is

work by gravity on system = 58.8 Newtons * 2.1 meters = 123.48 Joules.

The potential energy of the system therefore decreases by 123.48 Joules; the PE change is PE change = - 123.48 Joules.

Since there are no dissipative forces acting on the system, the KE change of the system must be equal and opposite to the PE change (`dKE + `dPE = 0 when no work is done by dissipative forces).

The KE change will therefore be

KE change = + 123.48 Joules.

If the system has initial velocity 3.4 m/s, then its original KE is

KE0 = 1/2 m v0^2 = .5 * 15 kg * ( 3.4 m/s^2) = 86.7 Joules.

As it moves 2.1 meters, the PE change of the system is - 123.48 Joules so the KE change is + 123.48 Joules, and the system ends up with

KEf = KE0 + `dKE = 86.7 J + 123.48 J = 210.18 Joules.

Since 1/2 m vf^2 = KEf, vf = `sqrt( 2 KEf / m), where m is the total mass of the system.
We thus find that

final velocity of system = vf = `sqrt(2 * 210.18 Joules / ( 15 kg)) = 5.29377 m/s.

This agrees with the final velocity found from analyzing the uniformly accelerated motion of the system.

In the presence of friction:

The forces acting on the system will now include the frictional force on the first mass, which is the product of the coefficient of friction .2 and the normal force.

Since the first mass is not accelerating in the vertical direction (i.e., is in vertical equilibrium) the sum of the normal force and gravitational force on this mass must be 0.

We therefore find that the normal force, being equal and opposite to the gravitational force, must have magnitude 9 kg * 9.8 m/s^2 = 88.2 Newtons.

It follows that the frictional force has magnitude

magnitude of frictional force = f = .2 * 88.2 Newtons = 17.64.

This force acts in the direction opposite to the motion of the system--i.e., in the negative direction when the system is moving in the positive direction and in the positive direction when the system is moving in the negative direction.

Gravitational potential energy will change by the same amount as before, by - 123.48 Joules.

However the change in kinetic energy will not the equal and opposite to the change in gravitational potential energy, because we now have work being done by the system against the frictional force.

To overcome the frictional force the system must exert a force equal and opposite to the frictional force, so that against friction the system does work

net work by system against friction = 17.64 Newtons * 2.1 meters = 37.044 Joules.

Since in general `dW + `dPE + `dKE = 0, we see that `dKE = -(`dPE + `dW) = -( - 123.48 J + 37.044 J) = 86.436 Joules.

A less formal and perhaps more intuitive way to put this is that gravity does work 123.48 Joules on the system and friction does work - 37.044 Joules, so that the net work done on the system is 123.48 J - 37.044 Joules = 86.436 Joules.

Now if the system starts from velocity 3.4 m/s, with kinetic energy 86.7 Joules, it will end up with kinetic energy KEfin = 86.7 J + 86.436 J = 173.136 Joules.

As before we set this equal to 1/2 m v^2 and solve for v to obtain final velocity

final velocity from 3.4 m/s = `sqrt( 2 * KE / m) = 4.804665 m/s.