yes that's what I go the second time i did it but i must have rounded wrong cause i got 27.
so that means the block will be at
block
|--------------------------------[<]---------------|
0 55 (30cm)...
yep so at what distance does it stop 0 being at the beginning of the friction surface and 85 being the end
i worked it out to be about 78cm on the fourth or fifth rebound..
you?
okay the picture didnt work but basically it has a block mass that is pushed into a spring and then it slides out onto a 85cm surface with friction coefficient 0.27 and at the other end it has a curved end so it will slide back.
http://www.facebook.com/profile.php?id=521962077&ref=profile&nctrct=1238731741972#/photo.php?pid=1687315&id=521962077&ref=mf
Here is a picture of the problem i drew. The end is curved so the block will return, but the question asks where the block will end up :S
okay i got the KE to be;
= 1/2 k x^2
KE = 2.25
then work as
W = 0.27 * 0.85
W= 0.2295 J
so.. KE = 2.25
you can say:
2.25 = 1/2 m v ^2
plug in the mass and find velocity as 5.028m/s when the spring returns to equilibrium?
then how do i find where it comes to rest? =S
Homework Statement
A 187 g block is launched by compressing a spring of constant k=200N/m a distance of 15 cm. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has coefficient of friction...