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?

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:

=

=

=

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i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!