Solving a Spring/Force Problem: Mass, Spring Constant, and Acceleration

[morrisj753]

Homework Statement

A mass m is resting at equilibrium suspended from a vertical spring of natural length L and spring constant k inside a box. The box begins accelerating upward with acceleration a. How much closer does the equilibrium position of the mass move to the bottom of the box?

Diagram: http://www.aapt.org/Programs/contests/upload/olympiad_2008_fnet_ma.pdf
Problem 17
(Answer: ma/k)

Homework Equations

F = ma
F = -kx

The Attempt at a Solution

Well, the forces acting are: the force of the spring (Fs), the weight of the object(Fg), and the force of the elevator (Fe).
Fe + fs - mg = ma
Fe - kx = m (a+g)
I thought this was correct but am not sure where to progress from here.

 

[TSny]

Hello.

Be careful when specifying the forces that act on the mass m. Which of the three forces that you listed does not act on m?

 

[SignaturePF]

Is the equation simply:
F_s = ma, where a is acceleration of box,
-kx = ma
x = ma/k

 

[morrisj753]

@Signature PF:
You might have made an error, because you forgot about the negative, it'd be x = - ma / k

@TSny:
Fe i suppose does not act directly on m, but the weight DOES directly affect the mass m.
The way the problem is worded, the weight overcame the force of the spring.
mg - (-kx) = ma
x = m (a-g) / k,
so I am still making an error somewhere

 

[SignaturePF]

Morris - the sign is insignificant in this case, all it represents is compression vs extension. In this case, the negative is understood to mean that the spring is compressed, so I omitted it in the answer.

 

[TSny]

Fe i suppose does not act directly on m, but the weight DOES directly affect the mass m.
The way the problem is worded, the weight overcame the force of the spring.
mg - (-kx) = ma
x = m (a-g) / k,
so I am still making an error somewhere

Right, the only two forces acting on m are the force of gravity and the spring force.

You will need to be careful with the signs. Since the box is accelerating upward, it might be good to take upward as the positive direction. So gravity exerts a force of magnitude mg downward while the spring force exerts a force of magnitude kx upward. When you fix the signs and solve for x, you will want to compare the result to the value of x when the box is not accelerating.

 

[SignaturePF]

When the box is not accelerating:
kx = mg
x = mg / k

When the box has an acceleration a:
kx - mg = ma
kx = mg + ma
x = m(g+a) / k
Subtracting the two yields:
mg + ma - mg / k
= ma / k
This is the desired answer.