Physics, 02.11.2020 17:00 mrmendrala
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?
Answers
Answer from: JS28boss
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:
= 
= 
The work done will be:
= 
= 
= 
Answer from: Quest
spare some change and give it to the poor
explanation:
plz : ((
Answer from: Quest
Answer8.0m;
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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...
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