[JCHARD]

a 17KG cat is shot out of a hole in the side of a charged plate when the voltage is suddenly turned up to 5000V. The cat has a charge of 2.377^16c. He lands in a 3kg crate on wheels that has a µ=0.1 for o.50 meters. The crate is then put to stop by a spring 700 N/M. how much did the spring compress?

[Curlyben]

What do YOU think ?
While we're happy to HELP we wont do all the work for you.
Show us what you have done and where you are having problems..

[ebaines]

Hint: energy principles....

[ebaines]

Follow up, based on the OP's request for further assistance:

Please post your follow up questions in this thread instead of using PMs. But here's a further hint: voltage is a measure of potential energy per unit charge. Hence the initial amount of PE in the system is equal to the voltage of the plate times the charge on the cat. At the conclusion when the spring is fully compressed that energy has been converted into work done by friction and work done to compress the spring. This assumes that the initial velocity of the cat is zero, hence no initial kinetic energy.

[JCHARD]

Thank you, but I still can't seem to get the equations and the algebraic work. This question is very misleading to my classmates and I, we have not been informed of these types of questions before.

[ebaines]

Initial energy imparted on the cat: , where Q = charge on the cat and V = voltage on the plate.

Work done by friction: remember that work = force times distance, and the friction force is where is the coefficient offriction and m is the mass of the cart+ crate. So .

Work done compressing the spring: , where k = spring constant in newtons/meter and x = the amount the spring is compressed.

At the time of max compression of the spring the velocity of cat + crate is zero, so you can ignore kinetic energy.

Set and solve for x.

By the way, in your original post you said the charge on the cat is "2.377^16c" - please clarify this. Did you mean 2.377x 10^16 C? If so that's a ridiculously high amount of charge.