Force

Using the formula below, where $u$ , $v$ , $a$ and $S$ are the initial velocity, final velocity and constant acceleration of the car, and the distance it covers respectively:

$\begin{aligned} v^2-u^3 & = 2aS \\ \implies a &=\frac {v^2-u^2}{2S} \\ & = \frac {\left(\frac{180\times 1000}{60 \times 60}\right)^2-0^2}{2\times 100} \\ &= 12.5 \text{ ms}^{-2} \end{aligned}$

Let the mass of the car be $m$ . Then by Newton Second Law of Motion, the force applied by the engine is $F=ma=$ $3000\times 12.5=$ $\boxed{37500}$ N.

Using that the total work is equal to the variation of kinectic energy:

$W=\Delta { E }_{ k }\rightarrow Fd=\frac { m{ v }_{ f }^{ 2 } }{ 2 } -\frac { m{ v }_{ i }^{ 2 } }{ 2 }$

Substituting the values:

$F*100=\frac { 3*1000*{ 180 }^{ 2 } }{ 2*{ 3.6 }^{ 2 } }$

Therefore,

$\boxed{F=37500}$

We have , u=0 v=180*5/18=50m/s s=100m m=3 tons = 3000 kg

Using third equation of motion we have. a=v^2 - u^2 / 2s = 12.5 m/s^2

Using formula F = MA ,we get , F = 3000*12.5 = 37500 N

Here v=180 * 5/18=50 m/s and m=3 tons=3*1000=3000 kg We know that S={(v+u)/2}t or 100={(50+0)/2}t or 100=25t so t=4 s,then a=(v-u)/t=50/4=25/2 m/s^2 now F=ma=3000 * 25/2=37500 N

v=180kmph=50mps u=0 s=100m m=3ton=3000kg

v^{2} - u^{2} = 2as

2500 - 0 = 2 a 100

a = 2500/200 = 12.5

F = ma = 3000*12.5 = 37500N