Work and Changes in Kinetic Energy Experiment

[KatlynEdwards]

Homework Statement

In experiment 1, two hands push identical blocks of mas m toward each other across a level frictionless surface with a constant force of magnitude F over the interval from t1 to t2.

In each experiment both blocks start from rest at time t1 on a level frictionless surface; both hands exert the same magnitude for f over the interval from t1 to t2; and the blocks do not run into each other during this interval. The system is defined as the blocks and any spring or rod that may be connecting them.

Experiment 2:
Same as above, except for the hands push in the same direction.
Is the net external work on the system of the two blocks greater than, less than, or equal to that of experiment 1?

Experiment 3:
Same as experiment 1, except for the clocks are connected by a spring. The spring is initially neither compressed nor stretched.
Is the net external work on the system of the blocks greater than, less than, or equal to that of experiment 1?

Experiment 4:
Same as experiment 1, except for the mass of one of the blocks is now m/2.
Is the net external work on the system of the blocks greater than, less than, or equal to that of experiment 1?

Experiment 5:
Same as experiment 1, except for the blocks are connected by a rigid rod.
Is the net external work on the system of the blocks greater than, less than, or equal to that of experiment 1?

Homework Equations

Total Energy = the Sum of all the individual energies (Kinetic and Potential)

Kinetic Energy = 1/2 * M * (V^2)
Potential Energy = M * Height * g

The Attempt at a Solution

Experiment 2:
I think it would be equal to the net external work on the system in experiment 1 because it's the same force on the same mass over the same interval. And since the blocks are moving in the same directions as the forces they're positive.

Experiment 3:
In this experiment the system is gaining potential energy from the spring, so I would say that it is greater than that of experiment 1.

Experiment 4:
I think this system has less external work on the system because the mass of the second block is 1/2 that of the block in experiment 1 - thus it's lost 1/4 of the energy.

Experiment 5:
I think the external work on this system is zero because the blocks don't move when force is applied. Thus this experiment has less than experiment 1.




Is this the correct line of thinking?

Thanks!

 

[collinsmark]

Is this the correct line of thinking?

Be careful! Work is defined as force times distance -- not force times time. (Technically, work is the dot product of force and displacement, so you have to consider the direction of the particular displacement relative to the particular force too [Edit: although I don't think it matters too much for this particular problem]).

In all the experiments, the magnitude of the forces is always the same. And the time intervals are also equal. So you really should be asking yourself which experiments cause the objects to move a greater or smaller distance [Edit: i.e. a greater or smaller distance within the time period that the forces are applied.]

 

[KatlynEdwards]

Okay, so then in experiment 1, the work is zero because the work on block one cancels out the work on block two due to the opposing directions.

In experiment 2 the work is greater than 1 because there are two positive works being added together.

In experiment three, the work is still zero like in experiment 1, but the system gains potential energy.

In experiment 4 the net work is greater because the negative work is half the magnitude of the positive work.

In experiment 5 the net work is zero because the distance is zero.

Is this correct?

 

[collinsmark]

Okay, so then in experiment 1, the work is zero because the work on block one cancels out the work on block two due to the opposing directions.

Sorry if I confused you earlier, but that's not quite right.

Force and displacement are both vectors. The work done by a given external force is the dot product between the individual force and the individual displacement of the object on which the force acts. Yes, the individual displacements are in opposite directions, but for this problem the corresponding individual forces are also in opposite directions.

And as it turns out in this particular problem, the direction of given force is always the same as the displacement of the corresponding object. In other words, in this particular problem, the given force and corresponding displacement are always in the same direction.

In experiment 2 the work is greater than 1 because there are two positive works being added together.

In experiment three, the work is still zero like in experiment 1, but the system gains potential energy.

Redo the above, considering the fact that the directions of the individual forces match their corresponding displacement vector directions.

And you need to think of distance! The magnitudes of the external forces are always the same in all the experiments. The only thing that really changes is how far objects are pushed.
The greater the distance that an object is pushed (with the given time-interval that the force is applied), the greater the work.

In experiment 4 the net work is greater because the negative work is half the magnitude of the positive work.

Again, there isn't any negative work involved in this particular problem. The force and corresponding displacement vectors are always parallel.

But think about it this way. If a constant force is applied to an object for a set amount of time, will the object travel farther if the object is less massive or more massive (ignore friction)?

In experiment 5 the net work is zero because the distance is zero.

Now that, I agree with!

 

[rwisz]

Yes, your line of thinking is correct. In experiment 2, the net external work would be equal to experiment 1 because the force and mass are the same and the blocks are moving in the same direction. In experiment 3, the net external work would be greater than experiment 1 because the spring is adding potential energy to the system. In experiment 4, the net external work would be less than experiment 1 because the mass of one block is reduced and therefore less force is needed to move it. And in experiment 5, the net external work would be zero because the blocks do not move and therefore no work is being done on the system.