- #1

Zyrn

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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]

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 this.

F=-k/x

SPE=(1/2)kx

KE=(1/2)mv

v

What is the spring's force constant?

x=0.03875m

F

k=((mg)/sinθ)/x=((5kg*9.8m/s

The program marked this wrong

What is the spring's potential energy now?

x=0.03875m+0.29m=0.32875m

SPE=(1/2)kx

The program also marked this wrong.

How far along the ramp beyond this release point does the rock slide?

v

x=v

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]

[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 this.

## Homework Equations

F=-k/x

SPE=(1/2)kx

^{2}KE=(1/2)mv

^{2}v

^{2}=v_{i}^{2}+2ax## The Attempt at a Solution

What is the spring's force constant?

x=0.03875m

F

_{g}=(mg)/sinθk=((mg)/sinθ)/x=((5kg*9.8m/s

^{2})/sin60°)/0.03875m≈1460 N/mThe program marked this wrong

What is the spring's potential energy now?

x=0.03875m+0.29m=0.32875m

SPE=(1/2)kx

^{2}=(1/2)(1460N/m)(0.32875m)^{2}≈78.9 JThe program also marked this wrong.

How far along the ramp beyond this release point does the rock slide?

v

^{2}=2*KE/m=2*78.9 J/5kg≈31.56(m/s)^{2}x=v

^{2}/(2(g/sinθ))=31.56(m/s)^{2}/(2(9.8m/s^{2}/sin60°))≈1.395mThe 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]

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