## anonymous 4 years ago A 1.2 x 10^4 kg truck is travelling south at 22 m/s. (a) what net force is required to bring the truck to a stop in 330m?

1. JamesJ

By the Work-Energy theorem, the amount of work done on the truck is equal to its change in energy. Work, W, is given by W = Fd where F is a force applied in the direction of a distance d. Now, this must be equal to the change in KE. See what to do now?

2. JamesJ

What is the change in KE for the truck?

3. anonymous

is it possible to use $v_{f}^{2} = v_{i}^{2} + 2ad$ to get the answer to this question? because that's what my friend told me to use. and then she told me to use$a= \left( v _{f}^{2} - v_{i}^{2}\right) / 2d$ but then i want to know how she got from the first equation to the second equation...

4. JamesJ

It is ... let me come back to it. Follow my logic for now. What is the expression for KE?

5. JamesJ

$KE = \frac{1}{2}mv^2$ where v is speed

6. JamesJ

Hence change in KE is $\Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$ By the Work-Energy theorem, this change in KE must be equal to the amount of work done on the truck. That work is given by $Work = Fd = mad$ because F = ma

7. JamesJ

Therefore, as we have by the Work-Energy Theorem: $Work = \Delta KE$ this means $Fd = \frac{1}{2}m (v_f^2 - v_i^2)$ You know that v_f = 0, therefore $Fd = - \frac{1}{2} m v_i^2$ You know d, m, and v_i. Hence you can solve for F.

8. JamesJ

What we've done here is how we derive the other equation your friend is referring to. By Newton's second law, F = ma, hence Work = Fd = mad thus $mad = \frac{1}{2} m (v_f^2 - v_i^2)$ Cancel m from both sides, multiply through by 2 and we have $2ad = v_f^2 - v_i^2$ From that, you could also solve for acceleration a, and therefore calculate F using F = ma. But the first derivation I've written above is more fundamental, and better reflects the Physics of the situation.

9. anonymous

Oh, that makes more sense. i got $a= -0.733$ and when i substitute that to $F=ma$ i get $F=(1.2x10^{2} kg)(-0.733)$ $F=-8800N$ F=8800 N [N] $F= 8.8 \times 10^{3}$ N [N]