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Assoli18 [71]
3 years ago
11

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
Physics
1 answer:
Nastasia [14]3 years ago
4 0

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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3 years ago
A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for
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The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>

Why?

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Have a nice day!

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