Finding Initial Separation on an Inclined Plane with a Compressed Spring

[f25274]

Homework Statement

A 3.00 kg block starts from rest and slides a distance d down a frictionless 30.0° incline. While sliding, it comes into contact with an unstressed spring of negligible mass. The mass slides an additional 0.219 m as it is brought momentarily to rest by compression of the spring (k = 400 N/m). Find the initial separation d between mass and spring.

Homework Equations

Kinetic energy: mv^2/2-mu^2/2
Potential energy of a spring: kx^2/2
ΔK=-ΔU
Δx=Vt+0.5at^2
V=Vo+at

The Attempt at a Solution

kx^2/2=mv^2/2
400(0.219^2)=3v^2
v=2.529 m/s
a=9.8sin(30°)
2.529=4.9t
t=0.516
x=2.45*0.516^2
x=0.653

The answer is wrong, haha. But that is what I did, help me?

 

[gneill]

The block is still moving downslope (and lower in height) as the spring is compressed. So there's some additional energy you need to account for.

 

[EkaterinaAvd]

I'd use a little different approach for this problem:
in both initial and final state the spring is at the rest, so
1) consider whether there was any work done by force other than gravitational force and spring elastic force
2) depending on with what you come up in (1), use energy conservation law or energy-work theorem

in your solution,
in the x=2.45*0.516^2, what is 2.45?
Otherwise, everything looks good. I would recommend to get symbolic answer first, and only after that to plug-in numbers because of two reasons:
1) it is easier to detect errors. When you see symbols, you know what they mean, when you see numbers - they are just numbers.
2) when you have symbolic answer, you can see that effect can be different depending on numbers and you understand Physics better

 

[f25274]

The 2.45 was a/2
Acceleration due to gravity was gsin(30)=4.9
I'll try to use symbols in my answer

Would it be kinetic energy(before compressing)+potential energy (spring)= gravitational energy?

 

[f25274]

I don't think there was work done by an outside force. There is no friction in this problem.

 

[EkaterinaAvd]

Ok, I see. Calculation would be correct if compession displacement were much smaller then initial height of the block. But 0.653 m and 0.219 m are comparable. So, read post from Gneill.