Conservation of Energy with Spring Mass System on Inclined Plane

[JakeHowarth]

Homework Statement

(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.

Homework 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

The Attempt at a Solution

I've been able to prove choice B correct 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 is twice as large as the potential energy of the mass under compression length x. Proving choice B correct.

However, I'm having trouble proving the other one's incorrect, especially choice C and choice D because I'm looking at the problem as two parts:

1. Potential energy of the mass compressed on the spring TO kinetic energy just before release by spring
2. 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 implies 2ho = h, which proves choice C.

So, I don't know if choice B or choice C is right. If someone can help me clarify this, I would greatly appreciate it. I've been at this for hours.

 

[lightgrav]

as you say, PE depends on the square of the compression (½kx^2)
what happens when you square of the new compression , "2x" ?
is the Work done, and the PE stored, 2x as big as before?
(not only is the max Force 2x as strong, but the distance is also 2x as far)

 

[JakeHowarth]

Okay, I'll recheck some things.

 

[lightgrav]

Yes, B is incorrect. The hint was that it is the SPRING which has elastic PE, not the box.
So it must be that the box goes higher, or leaves the spring faster ... (c,d,e)

 

[rwisz]

First, let's clarify what is meant by "twice as high" in choice C. It is referring to the final height of the box on the incline, not the height of the box above the starting point. So, in order to compare the final heights, we need to set the potential energy at the starting point (before the box is released) to be the same in both cases. This means that we need to take into account the initial potential energy of the compressed spring.

Under compression length x:

(Eq1) F = kx
(Eq3) Us = (1/2)(F)x = (1/2)(kx)(x) = (1/2)kx²
(Eq2) Uo = Us = (1/2)kx²
(Eq4) Ko = 0 (no initial kinetic energy)
(Eq5) U = mgh

Setting Uo = U gives us:

(1/2)kx² = mgh
h = kx²/(2mg)

Under compression length 2x:

(Eq1) F = 2kx
(Eq3) Us = (1/2)(F)x = (1/2)(2kx)(x) = kx²
(Eq2) Uo = Us = kx²
(Eq4) Ko = 0 (no initial kinetic energy)
(Eq5) U = mgh

Setting Uo = U gives us:

kx² = mgh
h = kx²/(mg)

So, the final height in the case of compression length 2x is double the final height in the case of compression length x. This proves choice C to be correct.

To prove choice D incorrect, we can use the same method as above and compare the final velocities of the box on the incline.

Under compression length x:

(Eq2) Uo = Us = (1/2)kx²
(Eq3, Eq4) (1/2)kx² = (1/2)mv²
v = sqrt(kx²/m)

Under compression length 2x:

(Eq2) Uo = Us = kx²
(Eq3, Eq4) kx² = (1/2)mv²
v = sqrt(2kx²/m)

So, the final velocity in the case of