# Hooke's Law/Work done on a spring system

1. Nov 18, 2013

### ZeGinger

1. The problem statement, all variables and given/known data

I've been trying to answer a Physics-based question, and I can't seem to correctly answer the problem correctly. it is a five-part problem, and since I cannot answer the first problem correctly, the other four cannot be answered.
It is...

The block in the figure below lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 35 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 3.2 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. Assume that the stopping point is reached.
(a) What is the position of the block?
(b) What is the work that has been done on the block by the applied force?
(c) What is the work that has been done on the block by the spring force?

I also need to find answers to these questions with regards to: 'during the block's displacement.'
(d) The block's position when its kinetic energy is maximum
(e) The value of that maximum kinetic energy.

2. Relevant equations

This particular problem utilizes Hooke's law and Work/Kinetic energy theorems. so far the only one I've used was Hooke's Law $F = kx$
I'm aware that I will need to use work equations to solve for b and c W = .5kxi2 - .5kxf2

As for d and e, I've not a clue how to answer.

3. The attempt at a solution

When solving for part a, I simply manipulated Hooke's Law to solve using the aforementioned values (3.2N and 35N/M). I ended up getting 0.0914 meters, but it is considered incorrect.
Am I missing something in the equation?

2. Nov 18, 2013

### Staff: Mentor

You left out the mass times acceleration in your equation. Even though the mass has stopped, its acceleration at that point is backwards towards its initial position. You need to set the work done by the force equal to the energy stored in the spring, since the kinetic energy at maximum extension is zero. You will find that you underestimated by final extension by a factor of 2. Basically, the mass overshoots the equilibrium location you calculated.

3. Nov 18, 2013

### ZeGinger

Hm, I'm not sure if I'm following. how would integrating Mass and Acceleration into the equation change the output of the problem? wouldn't the force value be all that's required...
or am I using mass and acceleration to calculated the fallback distance from the spring force?

4. Nov 18, 2013

### Staff: Mentor

The differential equation is:
$$m\frac{d^2x}{dt^2}=F-kx$$
Notice that the left hand side is not equal to zero (as you assumed). Solve the equation as a function of t, subject to the initial conditions x = 0 and dx/dt = 0 at time t = 0. Find the distance at which the velocity is again equal to zero. What do you get?