$k=8.1 \times 10^4N/m$

### Transcript for this Giancoli solution

We know that the size of the change in kinetic energy for this car. So we're going to go up to the value of change in kinetic energy which is one half 'mv' initial squared, the final speed is zero because the car comes to a stop. So we don’t need that term in this difference, this subtracting the final from the initial. We can just write down the initial kinetic energy because it loses all that it had to begin with. Is equals the change in potential energy which is going to be the potential energy gained by the spring. So one half 'kx' squared. So we have one half 'mvi' squared equals one half 'kx' squared.I'm putting these absolute value signs here just to say that negative signs don’t really matter as long as we keep in mind what's going on. What these are saying is that, the amount of change in kinetic energy equals the amount of the change in potential energy.

So we'll solve for 'k'. So dividing both sides by 'x' squared, and then noticing that one half is a common factor which cancels on both sides. Gives us 'k' is 'mvi' squared over 'x' squared. This is 1,200 kilograms times the speed which we'll have to change into meters per second. So we have 45 kilometers per hour divide by 3.6 to change into meters per second, square that and divide by2.2 meters squared. That’s ANSWER newtons per meters is the spring constant.

## Comments

Why can't we use the conservation of energy law here? Wext= Change in KE + Change in PE + Ethermal where we can take everything as the system so that no external work is done and Etherm is zero.

Hi hsumail, thanks for the question, and apologies for taking so long to get back. Conservation of energy is being used here, though it could be expressed as the work-energy theorem as well and get the same result. The idea is that the total energy before a process is the same as the total energy after the process, and the formulae surrounding this subject serve just to help account for all the different forms of energy before and after the processes. Conservation of energy problems are book-keeping exercises.

Now I'll get to your comment. You mentioned the work-energy theorem, so let's clarify that it's $W_{ext} = \Delta KE$, not the expression written in your comment. [edit: earlier comments about the Earth don't apply. I hadn't actually looked up the problem.] You could consider the car-spring as a single system, which is all well and good, but this isn't strategic for problem solving since your only conclusion would be that $W_{ext} = 0$. This problem is asking for the spring constant, and conservation of energy is the ticket to solving for it. Maybe you would prefer the following expression, rather than the one I wrote in the video:

$KE_1 + PE_1 + W_{NC} = KE_2 + PE_2$. $W_{NC}$ typically represents friction, which is zero in this case, so we plug different types of energy into this expression and get $\dfrac{1}{2}mv_i^2 + 0 + 0 = 0 + \dfrac{1}{2}kx^2$ where I've substituted specific expressions in the same order as the equation before that has more generic expressions. This will arrive at $\dfrac{1}{2}mv_i^2 = \dfrac{1}{2}kx^2$, which is the same as in the video.

Hope this helps,

Mr. Dychko

*Work-Energy Theorem