(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

(Figure attached to post)

A box of mass m is pressed against (but is not attached to) an ideal spring of force constant k, compressing the spring a distance x. After it is released, the box slides up a frictionless incline as shown in Fig. 7.5 and eventually stops. If we repeat this experiment but instead compress the spring a distance of 2x:

A)Just as it moves free of the spring, the box will have twice as much kinetic energy as before.

B)Just before it is released, the box has twice as much elastic potential energy as before.

C)The box will go up the incline twice as high as before.

D)Just as it moves free of the spring, the box will be traveling twice as fast as before.

E)Just as it moves free of the spring, the box will be traveling four times as fast as before.

The weight and the normal force from the surface are balanced, so the only force acting on the box is the work done in compressing the spring = kx.

2. Relevant equations

(Eq1) Hooke's Law: F = -kx

(Eq2) Conservation of Energy (no external work): Uo + Ko = U + K

(Eq3) Potential Energy of a Spring: (1/2)kx²

(Eq4) Kinetic Energy: (1/2)mv²

(Eq5) Potential Energy of Gravity: mgh

3. The attempt at a solution

I've been able to provechoice Bcorrect by doing the following:

Under compression length x:

(Eq1) F = kx

(Eq3) Us = (1/2)(F)x = (1/2)(kx)(x) = (1/2)kx²

Under compression length 2x:

(Eq1) F = 2kx

(Eq3) Us = (1/2)(F)x = (1/2)(2kx)(x) = kx²

thus, (1/2)kx² * 2 = kx², meaning the potential energy of the mass under compression length 2x istwiceas large as the potential energy of the mass under compression length x. Provingchoice Bcorrect.

However, I'm having trouble proving the other one's incorrect, especiallychoice Candchoice Dbecause I'm looking at the problem as two parts:

- Potential energy of the mass compressed on the spring TO kinetic energy just before release by spring
- Initial kinetic energy of just released mass TO final potential energy of mass up ramp

So,

Under compression length x:

(Eq2) Uo = K (no initial kinetic because mass is compressed on spring with no movement, no final potential because this is only up to mass release by spring)

(Eq3, Eq4) (1/2)kx² = (1/2)mv²

v² = kx²/m

Now we switch to part 2, where the mass has been just released from the spring (initial kinetic, no potential) to final position on ramp (final potential, no kinetic)

(Eq2) Ko = U

(Eq3, Eq5) (1/2)m(kx²/m) = mgh

(1/2)kx²/mg = h

Under compression length 2x:

(Eq2) Uo = K (no initial kinetic because mass is compressed on spring with no movement, no final potential because this is only up to mass release by spring)

(Eq3, Eq4) kx² = (1/2)mv²

v² = 2kx²/m

Now we switch to part 2, where the mass has been just released from the spring (initial kinetic, no potential) to final position on ramp (final potential, no kinetic)

(Eq2) Ko = U

(Eq3, Eq5) (1/2)m(2kx²/m) = mgh

kx²/mg = h

Here is the problem: With the height ho = (1/2)kx²/mg under compression length x and h = kx²/mg under compression length 2x, this implies2ho = h, which proveschoice C.

So, I don't know ifchoice Borchoice Cis right. If someone can help me clarify this, I would greatly appreciate it. I've been at this for hours.

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# Homework Help: Conservation of Energy with Spring Mass System on Inclined Plane

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