Question

Two identical 0.50-kg carts, each 0.10 m long, are at rest on a low-friction track and are connected by a spring that is initially at its relaxed length of 0.50 m and is of negligible inertia. You give the cart on the left a push to the right (that is, toward the other cart), exerting a constant 6.0-N force. You stop pushing at the instant when the cart has moved 0.60m . At this instant, the relative velocity of the two carts is zero and the spring is compressed to a length of 0.30 m. A locking mechanism keeps the spring compressed, and the two carts continue moving to the right. By what amount do you change the system's kinetic energy?

Answer 1

Answer:

The system's kinetic energy changes by 3.6 J

Explanation:

The given parameters are;

The number of cart = 2

The mass of each cart = 0.5kg

The initial length of the spring = 0.50 m

The final length of the spring =T0.3 m

The change in position of the first cart = 0.6 m

The energy given to the first cart = Work done by the force = Force × Displacement

The initial kinetic energy of the two cart moving together = Energy given to the first cart = 6.0 × 0.2 = 1.2J

The kinetic energy given to the two cart combined = The applied force × The total displacement of the two cart as they move together

The kinetic energy given to the two cart combined = 6.0 × (0.6 - 0.2)

The kinetic energy given to the two cart combined = 6.0 × 0.4 = 2.4 J

The total kinetic energy given to the two cart = 1.2 + 2.4 = 3.6 J

The total kinetic energy given to the two cart = 3.6 J

The system's kinetic energy changes by 3.6 J.