Set 3 Problem number 13

A mass of 9 kg on a level tabletop is attached by a light string to a mass of 5 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?

If the system is released from rest, how much work is done on it by the net force as it moves 2.5 meters under the influence of this force?
What therefore will be its kinetic energy after having moved the 2.5 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 .09.

Solution

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

In the absence of friction:

The net force on the system is the force exerted by gravity on the 5 kg mass.

The force of gravity on the 9 kg mass is equal and opposite to the normal force between the tabletop and this mass; the net force on this mass is therefore zero..

The net force on the system is therefore 5 kg * 9.8 m/s^2 = 49 Newtons.

The work on the system as it moves 2.5 meters will be the product `dW = F `ds = 49 Newtons * 2.5 meters = 122.5 Joules.

Since there are no dissipative forces this work will be equal to the increase in the kinetic energy of the system.

Since the system started from rest its initial kinetic energy was 0, so its final kinetic energy will be KEf = 122.5 Joules.

Its final velocity will therefore be v such that 1/2 m v^2 = KEf, so that

final velocity = v = `sqrt(2 KEf / m) = `sqrt( 2 * 122.5 Joules / ( 14 kg) ) = 4.1833 m/s.

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 49 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 = 49 Newtons * 2.5 meters = 122.5 Joules.

The potential energy of the system therefore decreases by 122.5 Joules; the PE change is PE change = - 122.5 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 = + 122.5 Joules.

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 .09 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 = .09 * 88.2 Newtons = 7.938001.

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.

As the system accelerates from rest it moves in the positive direction, and the net force on the system is therefore the sum of the forces acting in the directin of motion:

net force on system = Fnet = 49 - 7.938001 = 41.062 Newtons.

The work done by the net force on the system as is moves through a displacement of 2.5 meters is now

work done by net force = `dWnetForce = 41.062 Newtons * 2.5 meters = 102.655 Joules.

System kinetic energy, starting from rest, will thus be

system KE = 102.655 Joules.

Gravitational potential energy will change by the same amount as before, by - 122.5 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 = 7.938001 Newtons * 2.5 meters = 19.845 Joules.

Since in general `dW + `dPE + `dKE = 0, we see that `dKE = -(`dPE + `dW) = -( - 122.5 J + 19.845 J) = 102.655 Joules.

A less formal and perhaps more intuitive way to put this is that gravity does work 122.5 Joules on the system and friction does work - 19.845 Joules, so that the net work done on the system is 122.5 J - 19.845 Joules = 102.655 Joules.