Understanding Spring Force and Potential Energy on a Tilted Ramp

[Zyrn]

I figured out where I was going wrong, it was in the trig for getting the force down the ramp. mgsinθ, not mg/sinθ, although I still got the right answer for the last part using mg/sinθ.

[STRIKE]

Homework Statement

A spring is set to move along the surface of a frictionless ramp, which is tilted at an angle θ = 60 degrees away from horizontal. A rock of mass M = 5 kg is placed on a spring. As a result, the spring compresses, coming to rest a distance x = 3.875 cm along the ramp away from its original position.

What is the spring's force constant?

The rock is pushed down along the ramp an additional distance y = 29 cm. What is the spring's potential energy now?

The rock is now released, so that the spring flings it up along the ramp. The spring and rock lose contact when the rock reaches the spring's rest length. How far along the ramp beyond this release point does the rock slide? (You may assume that the ramp is large enough that the rock never reaches its top end).

The set-up looked like http://spiff.rit.edu/webwork2_course_files/phys216/tmp/gif/set17-prob4-./tilt_rock_spring.gif.

Homework Equations

F=-k/x
SPE=(1/2)kx²
KE=(1/2)mv²
v²=v_i²+2ax

The Attempt at a Solution

What is the spring's force constant?
x=0.03875m
F_g=(mg)/sinθ
k=((mg)/sinθ)/x=((5kg*9.8m/s²)/sin60°)/0.03875m≈1460 N/m
The program marked this wrong

What is the spring's potential energy now?
x=0.03875m+0.29m=0.32875m
SPE=(1/2)kx²=(1/2)(1460N/m)(0.32875m)²≈78.9 J
The program also marked this wrong.

How far along the ramp beyond this release point does the rock slide?
v²=2*KE/m=2*78.9 J/5kg≈31.56(m/s)²
x=v²/(2(g/sinθ))=31.56(m/s)²/(2(9.8m/s²/sin60°))≈1.395m
The program marked this correct.

Based on that I'm thinking that either the professor put in the incorrect answers for the first two, or I did something wrong that managed to sort itself out by the last question.[/STRIKE]

 

[NihalSh]

What is the spring's force constant?
x=0.03875m
F_g=(mg)/sinθ
k=((mg)/sinθ)/x=((5kg*9.8m/s²)/sin60°)/0.03875m≈1460 N/m
The program marked this wrong

Check $F_{g_{x}}$. you calculated it wrong.

$$F_{g_{x}}=-m.g.Sin(θ)$$

Edit: I guess you just figured it out!

 

[D H]

I figured out where I was going wrong, it was in the trig for getting the force down the ramp. mgsinθ, not mg/sinθ, although I still got the right answer for the last part using mg/sinθ.

Yep. You made the same mistake twice, and that resulted in the mistake canceling itself out. Every once in a while two wrongs *do* make aright. Don't count in it, however. Two wrongs more often than not compound the problem rather than canceling it out.