# Block, Ramp, Friction, and Spring

1. Oct 12, 2008

### r34racer01

A man pulls a block of mass m = 19 kg up an incline at a slow constant velocity for a distance of d = 6 m. The incline makes an angle q = 32° with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is µk = 0.2.

a) What is the work Wm done by the man?
I got Wm =782.31

At the top of the incline, the string breaks and the block, assumed to be at rest when the string breaks, slides down a distance d = 6 m before it reaches a frictionless horizontal surface. A spring is mounted horizontally on the frictionless surface with one end attached to a wall. The block hits the spring, compresses it a distance L = 0.4 m, then rebounds back from the spring, retraces its path along the horizontal surface, and climbs up the incline.

b) What is the speed v of the block when it first reaches the horizontal surface?

c) What is the spring constant k of the spring?

d) How far up the incline d1 does the block rebound?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 12, 2008

### Hootenanny

Staff Emeritus
You can attempt this question in one of two ways: energy conservation or forces analysis.

3. Oct 12, 2008

### r34racer01

Ok, well which do you recommend?

4. Oct 13, 2008

### Hootenanny

Staff Emeritus
Personally, I always find conservation of energy the most straight forward.

5. Oct 13, 2008

### r34racer01

Ok well some recommended I try v = sqrt 2(gh- ug cos 33), but that didn't work so now I'm lost again. How should I apply conservation of energy here?

6. Oct 13, 2008

### Hootenanny

Staff Emeritus
What does the principle of conservation of energy (COE) state? Can you use COE to write down an equation for part (b)?

7. Oct 13, 2008

### r34racer01

Ok so using COE I was able to figure out the the Wg+Wf = KE at the bottom and setting that equal to .5mv^2 I got v = 6.51.

Now I'm trying to solve for K when the spring is compressed. I know that at this point all the energy is in the spring so I thought KE = -k(x2^2 - x1^2)/2. So I did 403.04 = - k(0.6^2)/2 but that didn't work. Anyone know what I did wrong?

8. Oct 14, 2008

### Hootenanny

Staff Emeritus
Are you sure about that equation? How much energy does the block have at the top of the incline?