Block held against a spring going up a frictionless plane

[TaxiDriver113]

Homework Statement

A 3 kg block is held against a spring (k=580nt/m) at the bottom of a frictionless 25 degree incline. The spring is compressed 0.18m. How far along the incline does the block go when released.

Homework Equations

Usp=1/2kx^2

The Attempt at a Solution

So far the only thing I've been able to do was calculate the velocity which i got to be 1.65m/s if I did it correctly.

Usp= 1/2 * 580 * .18^2
= 9.396

then I did this equation but I didn't know exactly what h was that I was trying to calculate. I think i just ended up calculating how much the spring was compressed and I already knew that.

9.396 + (1/2)(3)(0) = (3)(9.8)(h) + (1/2)(3)(1.654^2)
9.396 = 29.4h + 4.103
h = .18

 

[TaxiDriver113]

did I not supply enough information?

 

[LowlyPion]

Welcome to PF.

The PE at the bottom should really give you your height shouldn't it?

 

[nrqed]

Homework Statement

A 3 kg block is held against a spring (k=580nt/m) at the bottom of a frictionless 25 degree incline. The spring is compressed 0.18m. How far along the incline does the block go when released.

Homework Equations

Usp=1/2kx^2

The Attempt at a Solution

So far the only thing I've been able to do was calculate the velocity which i got to be 1.65m/s if I did it correctly.

The speed at what position?

Actually, you don't need to find any speed to solve the problem. At the beginning and at the end of the motion (at the max height) the speed is zero.

Usp= 1/2 * 580 * .18^2
= 9.396

then I did this equation but I didn't know exactly what h was that I was trying to calculate. I think i just ended up calculating how much the spring was compressed and I already knew that.

9.396 + (1/2)(3)(0) = (3)(9.8)(h) + (1/2)(3)(1.654^2)
9.396 = 29.4h + 4.103
h = .18

You could simply do

9.396 J = 3 * 9.8 * h

find h and then use the angle to find the distance along the incline.