set force: small Δt | set force: large Δt

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In Illustration 8.1, we learned a change in momentum was due to a net applied force. What about kinetic energy? Well, it is also due to a net applied force, but in a different way. Recall that we talk about work as the amount of force in the direction of an object's displacement multiplied by the displacement. No displacement, no work. Work is positive if F and Δx (or dx) point in the same direction and negative if F and Δx (or dx) point in opposite directions. Restart.

Consider the force applied by the hand over a small Δt (this happens automatically at t = 1 s). Notice the change in momentum (position is given in meters and time is given in seconds). Initially the mass of the cart is 1 kg. Change the mass to 2 kg. Does the change in momentum differ? No!  But what does change is the final velocity; it is half of the velocity when the mass was 1 kg. The same force results in the same change in momentum in the same time interval.

So what happens to the kinetic energy? Does it remain the same upon a change in mass? No. Why not? Recall that the work that will be equal to the change in kinetic energy is related to the displacement the cart undergoes when the force is applied. Due to the larger mass, the cart does not accelerate as much and therefore does not move as far, so its kinetic energy is less.

Another way to represent this is in terms of the integral (the area) under a force cos(θ) vs. distance graph. This area is called the work that is the object's DKE. What can you say about the work received by the cart when its mass changes? Check the second box to find out. Again, it should be, and is, different.

What happens when instead of applying a small Δt, you apply a large Δt? There is a larger impulse because Δt is larger. There is also a larger change in the kinetic energy since Δx is larger as well.