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?

Question

jamesj · Answer

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?

jamesj · Answer

What is the change in KE for the truck?

anonymous · Answer

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...

jamesj · Answer

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

jamesj · Answer

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

jamesj · Answer

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

jamesj · Answer

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.

jamesj · Answer

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.

anonymous · Answer

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]