# Homework Help: Work-Energy Theorum: Spring potential energy vs Kinetic Energy

1. Oct 10, 2009

### Senjai

[SOLVED] Work-Energy Theorum: Spring potential energy vs Kinetic Energy

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

A 1350-kg car rolling on a horizontal surface has a speed v = 40 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.5 m. What is the spring constant of the spring? Ignore Friction and assume spring is mass-less.

2. Relevant equations
$$W = \Delta E$$
$$E_{pspring} = \frac{1}{2}(kx^2)$$
$$E_k = \frac{1}{2}(mv^2)$$

3. The attempt at a solution

First right off the bat, i converted 40 km/h to its m/s equivalent of aprox. 11.11 m/s

i state the law of conservation of energy: Energy before = Energy after

Therefore:

$$E_k = E_{pspring} \frac{1}{2}(mv^2) = \frac{1}{2}(kx^2)$$

then i isolate k

$$k = \frac{-mv^2}{x^2}$$

now heres the issue, is x negative? because the displacement is against the direction of motion?
and 2.5m = x, (-2.5)^2 gives me a answer of 4266 Nm
but -(2.5)^2 is entirely different.. This has been a long lasting math issue for me.

And what if x is positive?

i know k MUST be positive right?

Last edited: Oct 11, 2009
2. Oct 10, 2009

### rl.bhat

(2.5)^2 is correct. There is no negative energy in the nature.

3. Oct 10, 2009

### Delphi51

or in k = mv^2/x^2. No way you can get k negative!
The minus sign in F = -kx is supposed to help keep track of the fact that the force of the spring is opposite to the direction of stretch but it does seem to have a habit of getting in the way. k is ALWAYS positive.

4. Oct 11, 2009

### Senjai

Thanks, the negative sign on mv^2 was an algebra error... Thanks for the clarification guys!

5. May 18, 2010

### vaibhav1803

Your attempt is correct but you missed somethingthat for a spring if you take natural length as the datum, the force on change in length is given as:
$$\vec{F}= -k \vec{x}$$

and hence work done by a spring against external forces
$$W_{s}=\int\vec{F}\vec{.dx}$$
over the required limits

in our case the answer is
$$W_{s}=-\frac{kx^{2}}{2}$$
as
$$W_{s}=\Delta E$$
$$\Delta E=-\frac{mv^{2}}{2}$$
the change part was where you lost it all...the KE FELL TO ZERO. HENCE A NEGATIVE CHANGE.

Last edited: May 18, 2010