# Work and Energy Problem

1. Oct 14, 2012

### pxp004

An object with inertia m = 1.5 kg is connected to a spring with spring constant k = 250 N/m and equilibrium position at x0 = 5 cm (for x > x0, the force is directed toward the origin). The object is initially at rest at position x = 10 cm. You grab the object and move it slowly to position x = 20 cm. After you finish moving it, you continue to hold the object at rest at its new position. How much work did you do on the object? I know the main equations but i just done know what to do. Please Help!

2. Oct 15, 2012

### Simon Bridge

Welcome to PF;
You know the main equations, so a good first step is to write them down.

You need to know - conservation of energy, work-energy relation, potential energy stored in a spring.

3. Oct 15, 2012

### pxp004

could you help me get started. With what equation? and I'm kinda confused on the 3 distances given.

4. Oct 15, 2012

### HallsofIvy

Staff Emeritus
If it requires force kx to move a mass M as distance x, then the work necessary will be $(1/2)kx^2$. Is that the formula you needed?

5. Oct 15, 2012

### Simon Bridge

Lets see - the three distances are:
x0, the equilibrium position.
x1, the initial position
x2, the final position

what's the problem?

If the mass were at x0, how much force would the spring exert on it?

The mass was at rest at x1 > x0, how much force does did spring exert on it? What was the total force on it? What do you deduce from that?

You moved the mass from x1 to x2 > x1 ...

6. Oct 16, 2012

### pxp004

Did you guys get 1.25 J as the answer. And if so, why did it give me the mass of. And how does that not affect work done?

My thought process: work is the change in energy which in this case is the change in potential (atleast I think it is). So I used the equation 1/2 kx squared.

7. Oct 17, 2012

### Simon Bridge

Notice that the mass starts off at rest at a position that is not the equilibrium position?
What would you deduce from that information?