Finding Compression Distance of Spring

[AdkinsJr]

Homework Statement

An inclined plane with 20.0 degree angle has a spring with k=500 N/m at the bottom of the incline. A block of mass m=2.5kg is placed on the plane at a distance d=.300m from the spring. From this position, the block is projected downward toward the spring with speed v=0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest.

Homework Equations

(1)$$x(t)=v_ot+\frac{1}{2}at^2$$
(2) $$V(t)=v_o+gsin(\theta)t$$

I think that the potential energy stored in the spring will be equal to the kinetic energy of the block at the equilibrium point.

(3)$$\frac{1}{2}mv_f^2=\frac{1}{2}kx_f^2$$

The Attempt at a Solution

With the x-axis parallel to the incline, the acceleration is cause by the component $$F_{gx}=gsin(\theta)$$

I want to find the velocity using (2), so I need to find the time:

$$d=v_ot+\frac{1}{2}at^2$$

$$t=\frac{-v_o \pm \sqrt{v_o^2+2dgsin(\theta)}}{gsin(\theta)}$$

I find t=.102s. Puting this in the velocity function gave v(.102s)=1.09m/s. Finally, I put this into $$x_f=\sqrt{mv_o^2k^{-1}}$$ to obtain about .077 m. In the book they give .131m. I can't figure out where I went wrong.

 

[tiny-tim]

Hi AdkinsJr!

You don't need to find anything in between, just call the compression x, and use conservation of energy, PE_i + KE_i = PE_f + KE_f

(and don't forget the PE includes both the spring and gravity)