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

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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

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Ngọc Diệp 5 years 2021-08-27T07:21:25+00:00 1 Answers 105 views 0

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  1. thugiang
    0
    2021-08-27T07:23:23+00:00
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    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

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