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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 4.5-N force. You stop pushing at the instant when the cart has moved 0.55 m . 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?

Respuesta :

Answer:

The right answer will be "2.025 Joule".

Explanation:

According to the question,

The initial position of CM will be:

= 0.35 m

The final position of CM will be:

= 0.8 m

Now,

The CM moves to the right by:

= [tex]0.8-0.35[/tex]

= [tex]0.45 \ m[/tex]

The work done will be:

= [tex]F\times Displacement[/tex]

= [tex]4.5\times 0.45[/tex]

= [tex]2.025 \ Joule[/tex]

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