Question

Two 0.50-kg carts are pushed toward each other from starting positions at either end of a 6.0-mlow-friction track. Each cart is pushed with a force of 5.0 N , and that force is exerted for a distance of 1.0 m.
Part A

What is the work done on the two-cart system?

Part B

What is the change in kinetic energy of the system?

Part C

What is the kinetic energy of the center of mass of the system?

Answer 1

Answer:

A) W=10J

B) $\Delta K=10J$

C) $K_{CM}=0J$

Explanation:

Since the problem is in one dimension, we can consider that one cart (1) is pushed in the positive direction while the other (2) is pushed in the negative one, so the work done on the system will be:

$W=F_1.d_1+F_2.d_2=(5N)(1m)+(-5N)(-1m)=10J$

Using the work-energy theorem $W_{Net}=\Delta K$, and since all work done on the system is the one calculated before, we have $\Delta K=10J$

The net force on the system is $F_{Net}=0N$, so there is no acceleration of the center of mass and it's kinetic energy is the initial one, $K_{CM}=0J$.