Question

A 2,200 kg car moving at 18 m/s hits a barrier and comes to a stop. How much work is done to bring the car to a stop?3.6 x 105J3.6 x 105J4.2 x 105J4.2 x 105J

Answer 1

Answer:

$3.6 \times {10}^{5}J$

Explanation:

From the question, mass(m)=2200kg, unitial velocity(u)=18m/s and final velocity(v)=0m/s

We can calculate the work done to bring the car to a stop from the relation;

$W = F \times S........eqn(1)$,where

W=Work done

F=Force

S=distance

Also,

$F = m \times a............eqn(2)$

Putting eqn(2) into equn(3) we obatin

$W = m \times a \times S......eqn(3)$

From the equation of motion;

$a= \frac{v - u}{t}$

and

$S = (\frac{u + v}{2})t$

Substituting these into eqn(3), we obtain;

$W =m \times ( \frac{v - u}{t}) \times ( \frac{u + v}{2})t$

$\implies W=m \times ( v - u) \times (u + v)\times\frac{t}{t} \times \frac{1}{2}$

$\implies W=m \times ( v - u\times (u + v)\times \frac{1}{2}$

Substituting the values of m,u and v into the equation, we obtain.

$\implies W=2200 \times ( 0 - 18) (18+ 0)\times \frac{1}{2}$

Simplifying, we obtain;

$\implies W=1100 \times - 18 \times 18$

$\implies W= - 356400 = - 3.564 \times {10}^{5}$

NB: The negative sign indicates that the car decelerated to a stop.

Hence the Work done on the car is

$3.6 \times {10}^{5}J$