# Homework Help: Block down ramp colliding with spring

1. Oct 11, 2004

### TraceBusta

A 3.2-kg block is released 34m from a massless spring with force constant 178N/m that is fixed along a frictionless plane inclined at an angle of 27° above the horizontal. Find the maximum compression of the spring.

This is what I have tried so far.
equation (1) (1/2 mv^2) = (1/2 kx^2)

equation (2) (1/2 mv^2)=d*mg(sin theta)

i used equation (2) to solve for v.
then using that value of v i solved (1)

my answer for x=2.333 meters, however that is wrong

2. Oct 11, 2004

### Staff: Mentor

In calculating the gravitational PE, don't forget to include the distance the block falls in compressing the spring a distance X.

3. Oct 11, 2004

### TraceBusta

ok, i think i understand the concept of that, but I'm not sure how I would set that up in equation form. Because gravity works in the y-component, would i have to use 2 equations. 1 to include for the y-distance the spring compresses and another to solve for the total compression in the sin (theta) component of the spring? If that was the case then I dont know how to set those up because i am trying to solve for maximum compression and in the first equation i woul dneed to know that.

4. Oct 11, 2004

### Staff: Mentor

Think of it this way: How far does the block slide? d = 34m + x. (x = how far the spring compresses.) Use energy conservation: Intial energy (gravitational PE) = final energy (spring PE). You'll get a quadratic equation.

Hint: measure gravitational PE from the lowest point.

5. Oct 11, 2004

### TraceBusta

i set up the problem like this
U=mg(h+x sin (theta))
Uspring=1/2 k x^2

so mg(h+x sin (theta)) = 1/2 kx^2?

solving for x, x1=1.1331, x2=-1.0965
those are both wrong.

6. Oct 11, 2004

### Staff: Mentor

Ugrav = mgd sin(theta), where d = 34m +x