- #1
Selophane
- 7
- 0
A moving 4.20 kg block collides with a horizontal spring whose spring constant is 270 N/m. The block compresses the spring a maximum distance of 4.00 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. What is the work done by the spring in bringing the block to rest?
When I first tried this I assumed it was simply:
W = 0.5kx^2 = (0.5)(270)(0.04)^2 = 0.216 J
However the answer is not matching with the correct one, not sure where i went wrong on such a simple problem... sorry if this is too easy, which I'm sure it is...
btw, there is a 2nd and 3rd part of the question
2nd part:
How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?
I got this one right, simply: W = Ffs = (4.2)(9.81)(0.25)(0.04) = 0.412 J
finally, part 3:
What is the speed of the block when it hits the spring?
now i tried simply using (work of friction) = 0.5mv^2 but am not getting correct answer?
Thanks,
Chris
When I first tried this I assumed it was simply:
W = 0.5kx^2 = (0.5)(270)(0.04)^2 = 0.216 J
However the answer is not matching with the correct one, not sure where i went wrong on such a simple problem... sorry if this is too easy, which I'm sure it is...
btw, there is a 2nd and 3rd part of the question
2nd part:
How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?
I got this one right, simply: W = Ffs = (4.2)(9.81)(0.25)(0.04) = 0.412 J
finally, part 3:
What is the speed of the block when it hits the spring?
now i tried simply using (work of friction) = 0.5mv^2 but am not getting correct answer?
Thanks,
Chris