A light horizontal spring has a constant of 135N/m.?

Question

A 2.18kg block is pressed against one of the ends of the block of the spring, which compresses the spring 0.101m. After it is released, the block moves 0.287m to the rightbefore it comes to a rest. Acceleration of gravity is 9.8m/s^s. What is the coefficient of kinetic friction between thesurface and the bloce?

Answer 1

There's a simpler way than that suggested by the first answer. As the 1st answer says, the potential energy is kx^2/2 and equals the kinetic energy at release. You don't have to know the release velocity and compute deceleration. We know that the kinetic energy is dissipated by 0.287 meters of friction force, so friction force = (kx^2/2)/0.287. The normal force is the weight mg, so the coefficient of friction = friction force/mg.
EDIT: I suppose it's possible that there is a question about the total movement of the block, depending on how you interpret "after it is released". I take that to mean after the person pressing the block against the spring lets go, which means the total travel is 0.287 m. Other answerers apparently interpret it as after the block separates from the spring, which adds another 0.101 m to the travel. Either way, it's the method of solution that is important. A minor quibble with answer 1: d is not just the distance the block slowed down, it's the total distance the block travels. This includes the distance it is pushed by the spring, unless one assumes the friction only occurs during coasting, which isn't implied in the problem statement. Which brings up another issue: with friction during the push, the block's kinetic energy never equals the spring's potential energy, so the velocity at separation is not as answer 1 and my answer state. But (again) using simple equivalence of potential and frictional energy, we don't need to know it anyway.

Answer 2

you know the amount by which the spring is compressed, which is .101m, and the spring constant, which is 135 N/m, so just use Hooke's law:
F=kx,
where k is the spring constant and x is the distance that the spring is compressed. this much force is used to compress the spring, so by newton's 3rd law, the spring exerts this much force on the block. when the block stopps pressing, the spring pushes the block back through the distance x, so it does work on the block equal to the integral of kx with respect to x, or kx^2/2. this work is also equal to the change in kinetic energy of the block. since the block started out at rest, its final kinetic energy, mv^2/2, is equal to kx^2/2. you now know the kinetic energy when it's released, and you know that it stops .287 m away, so use the equation:
v^2=v0^2+2ad,
where v is the final velocity (here, it's 0), v0 is the release velocity, and d is the distance through which it slowed down. you can use this to solve for the deceleration of the block. multiply this by the mass and you get the force of friction. divide the friction force by the magnitude of the normal force (here, it's equal to the weight of the object, mg) and you get the kinetic friction coefficient.

Answer 3

Initially you have 1/2kx^2 worth of potential energy or

W= ½ (135) (.101) ^2 = 0.689J

If you set this equal to the work done by friction to stop the block you get
[W=Fd= (uN) d= u (mg) d] and you can solve for:

u = W/ (mgd)
(with d equal to 0.388m)

Answer 4

http://en.wikipedia.org/wiki/Coefficient_of_friction