# Potential Energy and Conservation of Energy

1. Nov 8, 2015

### fromhuk

1. The problem statement, all variables and given/known data
A 700g block is released from rest at height Ho(initial) above a vertical spring with spring constant k=400 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 19.0 cm.
How much work is done
A) by the block on the spring?
B) by the spring on the block?
C) What is the value of Ho?
D) If the block were released from height 2ho above the spring, what would be the maximum compression of the spring?

2. Relevant equations
W= ΔK

Conservation of Energy:
Eo= Ef
Ug(o) + Us(o) + Ko = Ug(f) + Us(f) +Kf

3. The attempt at a solution

mgho + 0 (because the spring is not compressed at the start) + 0 (because the initial velocity is 0) = mg(-0.19m) (setting the top of the spring at y=0) + (kx^2)/2 + 0 (because the final instantaneous velocity is 0)

ΣW = Kf - Ko = Ug(o) - Ug(f) - Us(f) = mgho - mg(-0.19m) - 7.22J

I am mainly struggling with the conceptual part... for (a), the work done is only the compression of the spring, not the total work, right? So, it would be 7.22J. And (b) would be negative because the block is exerting force on the spring, and the spring is experiencing work that it can later convert into its own form of work?

For (c), solving Ug(o) + 0 + Ko = Ug(f) + Us(f) +Kf for ho:

W = mgho - 7.22J - mg(-0.19m) = 0
ho + (0.19m) = 7.22J/6.86N
ho = 0.86meters

For (d):

W = 2mgho - (k(x^2)/2) - mg(-0.19m) = 0
x^2 = 2(13.10J)/400N/m
x = 0.26meters

If someone could help explain work vs. forces vs. energy as it relates to this problem it'd be much appreciated!

2. Nov 8, 2015

### Andrew Mason

Welcome to PF fromhuk!

H0 is the initial distance of the block from the top of the uncompressed spring. The total distance the block falls, therefore is ________.

When the spring is compressed to 19 cm from its original state, is there any kinetic energy in the system (spring + block)? What is the potential energy stored in the spring at that point? What is the change in potential energy of the block? Set up an equation that relates the energies and you will be able to solve this.

Work is defined as: $W = \int \vec{F}\cdot d\vec{s}$. The block applies a downward force on the spring as it moves downward, so the directions of the force and displacement are the same. Therefore, the block does positive work on the spring. The spring, however, applies an upward force on the block as the block moves downward, so the force and the displacement through which that force acts have opposite signs. This means the spring does negative work on the block.

AM

3. Nov 8, 2015

### fromhuk

ho + 0.19m, however, I set the top of the spring as y=0, so those extra 0.19m do not have to be included in the gravitational potential energy, right?

I don't see why there'd be kinetic energy in the system when the spring is compressed- the block is not moving and the spring has potential energy Us: 7.22J, what am I missing? The change of potential energy in the block would be Kf - Ki + Uf - Uo = -Us; -7.22J.

PS. Thanks for the help!

4. Nov 8, 2015

### vela

Staff Emeritus
Right. The block exerts a force on the spring, which compresses the spring. The work done by the block is stored as potential energy by the spring.

Newton's third law says if the block exerts a force on the spring, the spring exerts an equal and opposite force on the block, so the work done by the block on the spring and the work done by spring on the block will always be related by a minus sign.

That's what I get too.

Why do you have mg(-0.19 m) in there? Is that still relevant when the block is dropped from twice the height above the spring?

You should stick with one variable, either x or y, for the vertical direction. You say the top of the spring is at y=0, but then you describe the compression of the spring as x. Pick one and stick with it.

5. Nov 8, 2015

### fromhuk

Thanks! You are right, of course. I have it down with "y"s rather than x, and that 0.19m as a y instead.
.

I did 2mgho - (ky^2)/2 - mgy
Solving for y:
y=0.23meters

I'm not sure if I messed up the symbols though... should I be adding the final potential forces instead of subtracting (because we know y is negative)? So it should rather be y=-0.26m?

6. Nov 8, 2015

### Andrew Mason

That is not an equation. You should explain your reasoning. Where does the 2 in 2mgho come from? y is the compression distance, which you are given (.19 m).

If you meant: $mg(h_0 + y) = \frac{1}{2}ky^2$ that would be correct. I get .86 m for h0

AM

7. Nov 8, 2015

### vela

Staff Emeritus
It depends on how you define $y$. If you define it to be the distance the spring is compressed and hence assume that $y>0$, the final gravitational potential energy would be $mg(-y)$, and you'd choose the positive root of the quadratic. If instead you take $y$ to be the displacement of the block, you'd use $mgy$, and you'd choose the negative root because the other root would be a spurious solution corresponding to a point above the spring.