# Homework Help: Help with Spring Question

1. Jan 24, 2012

### piggyxchu

Help with Spring Question!!

A toy contains a spring of spring constant 10 N/m. The spring is compressed 5.0 cm from its equilibrium position and can then released to propel a 6.0 gram marble.

a. Calculate the acceleration of the marble when the spring has expanded to 3.0 cm from the spring equilibrium position.

1. The problem statement, all variables and given/known data
k=10N/m
x=5cm=0.05m
mass=6g=0.006kg

2. Relevant equations
$\frac{1}{2}$k(x)$^{2}$

$\frac{1}{2}$mv$_{0}$+$\frac{1}{2}$k(x)$^{2}$ =$\frac{1}{2}$mv$_{f}$+$\frac{1}{2}$k(x)$^{2}$

3. The attempt at a solution
$\frac{1}{2}$(k)(x)$^{2}$ =$\frac{1}{2}$(m)v$^{2}$

$\frac{1}{2}$(10N/m)(0.04)$^{2}$ =$\frac{1}{2}$(0.006kg)v$^{2}$
v=5.77m/s

V=$\frac{Δx}{Δt}$
Δt=$\frac{0.08m}{5.77m/s}$=0.0139s

a=Δv/Δt=415.1m/s^2

b.The marble sticks to the spring for 1.0 cm AFTER it has passed its equilibrium position, and then leaves the gun. Calculate the speed of the bullet as it leaves the gun barrel

$\frac{1}{2}$(10N/m)(0.04)$^{2}$ =$\frac{1}{2}$(0.006kg)v$^{2}$
v$^{2}$= $\frac{0.2N}{0.003kg}$
v=√$\frac{0.2N}{0.003kg}$=8.16m/s

Last edited: Jan 24, 2012
2. Jan 24, 2012

### cepheid

Staff Emeritus
Re: Help with Spring Question!!

Welcome to PF!

Your approach to the first part of the problem is wrong, because the acceleration of the marble is NOT constant (because the force is not constant -- it varies with the position of the spring). The equation a = Δv/Δt is ONLY valid for constant accelerations.

Instead, to figure out the acceleration, use Newton's second law, and the fact that you know the force acting on the marble. The only force acting on the marble is the spring force, -kx, where x is the displacement from the equilibrium postion. Hence, Newton's 2nd law says that:

-kx = ma

Solve for a at the displacement, x, in question.

3. Jan 24, 2012

### Staff: Mentor

Re: Help with Spring Question!!

Redo that calculation.

V = Δx/Δt only applies for constant velocity motion (or average quantities). And why is your displacement set equal to 0.08m?

Hint: For this problem it may be easier to find the acceleration using Newton's 2nd law than by using energy methods.

Edit: Looks like cepheid has beaten me to the punch; you're in good hands.

4. Jan 24, 2012

### piggyxchu

Re: Help with Spring Question!!

I've set my displacement equal to 0.08m because I thought the displacement would be the compression of 5cm+ expansion of 3cm.....

So... Second Attempt.. :)
-kx=ma
-(10N/m)(0.03m)=(0.006kg)a
a=$\frac{-(10N/m)(0.03m)}{(0.006kg)}$
a=-50m/s^2

Why is acceleration negative? I understand that the negative sign comes from the equation... since the displacement here is an expansion of 3cm... does that mean I should be using -0.03m instead?

5. Jan 24, 2012

### Staff: Mentor

Re: Help with Spring Question!!

I see. The way I interpret the statement is that the spring was initially compressed to 5cm and allowed to decompress to 3cm for a displacement of 2cm.

The negative sign in Hooke's law just means that the restoring force is always opposite to the displacement. (I often ignore the sign when computing the magnitude of the force, then apply the correct sign to the final answer.)

Whether the sign of your acceleration is negative depends on your sign convention. The way I would do it is this. Define the direction that the marble is being shot to be positive. Thus the displacement from equilibrium would be -0.03 m and the acceleration would be positive (since the force is pushing the marble in the positive direction).

6. Jan 25, 2012

### cepheid

Staff Emeritus
Re: Help with Spring Question!!

Because of the part in bold above, I interpret it differently from you. Taking compression to be the negative x-direction and expansion the positive, I interpret the statement as saying that the spring is initially at x = -5 cm when it is released, and then allowed to expand to x = +3 cm, which is the point at which we want to evaluate the acceleration. So I think that the attempt in post #4 is correct.

EDIT: Under this interpretation, it also makes sense that the acceleration is negative, since the spring has crossed x = 0, is now stretched rather than compressed, and hence is beginning to pull back on the marble slightly.

7. Jan 25, 2012

### Staff: Mentor

Re: Help with Spring Question!!

Since the marble is being propelled by the spring and is not attached to it (except for sticking a bit in part b), I interpret it differently. And in light of part b it seems clear that in part a the marble has not yet passed equilibrium.

8. Jan 25, 2012

### cepheid

Staff Emeritus
Yeah, that is an issue. And since there is no mention of sticking (yet), the marble would be released from the spring at x = 0.

I see now the ambiguity in the wording. In the phrase "3 cm FROM the equilibrium position," the word "from" means "before" rather than "after." So you are right. I revise my interpretation to be that the marble moves from x = -5 cm to x = -3 cm. The acceleration is positive throughout. Sorry for any confusion.

9. Jan 28, 2012

### piggyxchu

Re: Help with Spring Question!!

sorry, that was a typo.. it was meant to be 0.02m instead of 0.04 which will result a velocity of 5.77m/s

the compression of the spring is 5 cm FROM it's equilibrium (-5) and the expansion of 3 cm is from the point of the 5 cm compression... Which will result a displacement of -2 cm (compression of 2 cm) from the equilibrium?

In this case, shouldn't i use -0.02m in the calculation of my acceleration instead?

-kx=ma
-(10N/m)(-0.02m)=(0.006kg)a

a=$\frac{−(10N/m)(-0.02m)}{(0.006kg)}$
a=33.33m/s^2

The wording of this question still throws me off.

On a side remark... the acceleration I've got from using -0.02m as the displacement of the spring is the same as the velocity squared? is this suppose to happen? :uhh:

10. Jan 28, 2012

### Staff: Mentor

Re: Help with Spring Question!!

The spring goes from x1 = -5cm to x2 = -3cm.
No. You want the displacement from the equilibrium position (x = 0). Note that the fact that the spring is initially compressed to x = -5cm is irrelevant for part a.

(The source of confusion is mixing up the meaning of 'displacement' as used in Hooke's law with the usual meaning of 'displacement'. In Hooke's law, displacement is always from equilibrium, whereas usually displacement means Δx.)

11. Jan 28, 2012

### piggyxchu

Re: Help with Spring Question!!

Okay :) So for part a. I should be using -0.03m as the displacement..as it is -3 cm FROM the equilibrium.

For part b, does that mean I should use 0.01m instead of 0.04m as it is 1m after it PASSED the equilibrium? Which results a speed of 4.08m/s..

Or is it compression of 5cm + expansion of 1cm=6 cm?

12. Jan 28, 2012

### Staff: Mentor

Re: Help with Spring Question!!

Right.
Yes, the position in part b is where x = 1 cm.
How did you arrive at that?
Nope. (That's the same confusion that I pointed out before.)

13. Jan 28, 2012

### piggyxchu

Re: Help with Spring Question!!

I've used the same formula as I did above..
$\frac{1}{2}$(k)(x)$^{2}$ =$\frac{1}{2}$(m)v$^{2}$

$\frac{1}{2}$(10N/m)(0.01m)$^{2}$ =$\frac{1}{2}$(0.006kg)v$^{2}$
v=1.29m/s

sorry... i forgot to square my 0.01m. I did my calculations again

14. Jan 28, 2012

### Staff: Mentor

Re: Help with Spring Question!!

You need to use conservation of total mechanical energy: Spring PE + KE.

15. Jan 28, 2012

### cepheid

Staff Emeritus
Re: Help with Spring Question!!

At the risk of hindering more than helping (again):

The total energy of the system is equal to the initial energy stored in the spring, which was the energy to compress it by 5 cm.

As the the spring expands back to the equilibrium (uncompressed, unstretched) position at x = 0, how much of this energy has been converted to kinetic energy? Ans: all of it.

Okay what about once it has stretched out to 1 cm? How much of the total energy is kinetic then?
Ans: slightly less than all of it, since some of it went back into stretching the spring.

This is the same method (cons. of energy) that doc al is advocating. I'm just walking you through it conceptually/in words.