Question

A 13kg box slides 4.0m down the frictionless ramp shown in the figure , then collides with a spring whose spring constant is 170 N/m.

Figure : http://i.imgur.com/0M7us.png

What is the maximum compression of the spring?
At what compression of the spring does the box have its maximum speed?

I'm stuck on the first question.

I thought I use mgh = (1/2)kx^2?
so, (13kg)(9.8m/s^2)(4sin(30)) = (1/2)(170N/m)x^2.

x = √(2mgh / k) = 1.7m?
But it's telling me this is wrong
Any help please?

Answer 1

I'm trying this process, but it doesn't seem to be working. I come up with x=\[\sqrt{2(13)(9.8)(\sin30^{o})(4)/170}\]
Which SHOULD be right, right?

Answer 2

i think

W = 1/2 k dx^2
F s = 1/2 k dx^2
(mg sin theta) s = 1/2 kdx^2
(mg sin theta)s = 1/2 kdx^2

dx = sqrt( (mg sin theta) s/(0.5 k))
dx = sqrt(2(mg sin theta)/k)

Answer 3

That's exactly what I'm doing! :(

Answer 4

dx = 0.865 m., right??

Answer 5

I'm getting 1.7

Answer 6

So whats the real answer?? thats multiple choice??

Answer 7

No, it's an online program. I WISH it was multiple choice. I mean that I'm calculating the answer as 1.7, and it's telling me I'm wrong.

Answer 8

And your method isn't working either! D:

Answer 9

Aha! I found it! It's the s. S equals how far the block slides PLUS the compression! Victory! :D Thanks for helping me think through, Gerry.

Answer 10

congrats @ChemicallyWrit