# Best solution? for work done on a spring.

1. Jun 11, 2013

### Tyrannosaurus_

1. The problem statement, all variables and given/known data
GIVEN: Force constant (k) = 170N/m. Displaced 10cm (0.1m) from equilibrium position.
REQUIRED: Find the work done on the spring.

2. Relevant equations
Solution 1 uses: W = ΔE, & (elastic energy} Ee = 0.5(k)x2, where k is the constant, and x is displacement from equilibrium position.

Solution 2 uses: W = F(Δd), & (elastic/spring force} Fe = kx, where k is the constant, and x is the displacement from the equilibrium position.

3. The attempt at a solution
Please evaluate these solutions & choose the best. Please tell me why. Please correct my misunderstandings. Thanks!

SOLUTION 1.
w = ΔE
w = Eefinal - Eeinitial
w = 0.5(k)xfinal2 - 0 {because initial the spring is at the eq. position.)
w = 0.5(170N/m)(0.1m)2
w = 0.85J

SOLUTION 2.
w = F(Δd)
w = Fe(Δd)
w = kx(Δd)
w = (170N/m)(0.1m)(0.1m)
w = 1.7J

Which solution is better, and why? (It's obvious why the values are different.)
Please help me understand why/where these approaches are best. (I understand calculus, but I need to understand these concepts, before I use calculus to evaluate.)

2. Jun 11, 2013

### Staff: Mentor

The second method is wrong, F is not constant during the displacement.

3. Jun 11, 2013

### Tyrannosaurus_

Thanks very much! ! !

Is there a resource you can point me to that explains why the force is not constant? Or, can you tell me why?

I'm not trying to be difficult, I just want to understand. I really want to be an engineer.

4. Jun 11, 2013

### Staff: Mentor

Just check how you calculated the force at 10cm excitation. What is the force at an excitation of 5cm?

5. Jun 11, 2013

### Saketh#

(elastic/spring force} Fe = kx, displacement x varies.Hence force is not constant