Question

A 1.60 kg block collides with a horizontal weightless spring of force constant 4.60 N/m. The block compresses the spring 4.00 m from the rest position. Assuming that the coefficient of kinetic friction between the block and the horizontal surface is 0.280 , what was the speed of the block (in meters/second) at the instant of collision?

The coefficient of friction is really throwing me off here. Do I use the spring constant?

I was using h=1/2gt^2 and using that to find t, then finding velocity in the horizontal direction, and then finding the vertical velocity.
My answer was wrong in th

Answer 1

You can solve this problem using conservation of energy. Keep in mind as well that everything is occurring in the horizontal direction in this problem.

At the moment the block hits the spring, it has some amount of kinetic energy, KE. As the block compresses the spring and eventually comes to a stop, part of KE is converted into elastic potential energy, PE, and some is dissipated as heat due to friction. The heat produced due to friction is in fact the work done by friction, W. In other words:

\[KE=PE+W\]
Do you think you could solve this problem now?

Answer 2

so....sorry if this is dumb, but is the equation going to be

\[\frac{ 1 }{ 2 }mv ^{2} = \frac{ 1 }{ 2 }kx ^{2} umg\]

with u being the coefficent of friction?

Answer 3

Not a dumb question! You almost got it:

\[{1 \over 2}mv^2={1 \over 2}kx^2+\mu mgx\]
Remember, W=Fd. μmg is the FORCE of friction, but you need to multiply the force by the distance over which the force acts, in this case the compression of the spring (x), in order to find the WORK.

Now you can solve :)

Answer 4

Yay! My final answer was 8.24 which is correct! Thank you so much!

Answer 5

No problem - glad I could help!