# Calculating a Spring Constant

1. Apr 30, 2010

### danvalen1

1. The problem statement, all variables and given/known data
A man that has a mass of 100kg stands on a platform 10m high. Right next to the platform is a spring of the same height. When the man steps off the platform and steps onto the spring, the spring compresses to 3m in height. What is the work done on the spring? What is the spring constant?

2. Relevant equations
w=(1/2)Fd
F=ma
w=(1/2)kD2

3. The attempt at a solution
w=(1/2)(100kgx10m/s/s)(10m-3m)=3500J

3500J=(1/2)k(7m)2
k=142.9 N/m

Am I using the correct equations and if so, am I getting the right answer?

Last edited: Apr 30, 2010
2. Apr 30, 2010

### Filip Larsen

Welcome to PF!

You seem to have used a valid approach to solve the problem, but the equation for work done by gravity on the man is not quite right and, not knowing what precision is required, I would probably expect a more precise value of g to be used.

3. Apr 30, 2010

### danvalen1

Using 9.8 m/s2:
w=(1/2)(F)(d)
w=(1/2)(100kg x 9.8 m/s2)(10m-3m)
w=3430J

I'm using w=(1/2)Fd because it is derived from w=(1/2)kd2 since k*d = F.

3430J=(1/2)k(7)2
k=140 N/m

4. Apr 30, 2010

### collinsmark

As Filip Larsen points out, you are not using the correct formula for the work done by gravity.

The formula for the force on a spring,

$$F = kd$$

is correct.

The formula for the potential energy of a spring, or work done on a spring,

$$W = \frac{1}{2}kd^2$$

is also correct. Let me show you where this comes from. By the definition of,

$$W = \int \vec F \cdot d \vec s$$

we have as applied to a spring (using the variable x instead of d, for the compression distance),

$$W = \int _0 ^x \vec F(x') \cdot d \vec x'$$

$$= \int _0 ^x kx dx,$$

Thus

$$W = \frac{1}{2} k x^2.$$

Or if you use d for the compression distance,

$$W = \frac{1}{2} k d^2.$$

But,

$$W \ne \frac{1}{2}Fd.$$

It doesn't "work" that way.

For gravity, assuming a constant acceleration, thus constant force over h,

$$W = \int \vec F \cdot d \vec s$$

becomes,

$$W = \int _0 ^h \vec F \cdot d \vec z$$

$$= \int _0 ^h mg dz.$$

But here (unlike the spring case), m and g (and F, for that matter) are constant over z, and can be pulled out from under the integral.

$$W = mg \int _0 ^h dz$$

Which becomes,

$$W = mgh$$

Or if you'd rather,

$$W = F _g h$$

where Fg is the force due to gravity (assuming constant acceleration).

The major difference between gravity and a spring, is that in the case of gravity, the force is constant. In the case of a spring, the force is proportional to the amount of compression. So they each have correspondingly different equations for their work functions (or potential energy functions, if you interpret it that way).

5. May 1, 2010

### danvalen1

Now I get why I can't use that. Thank you very much!

Last edited: May 1, 2010