Question

A car of mass 3*10^3 kg travelling at 54 km/h in straight road is brought to rest in a distance of 100 m by applying breaks. Calculate the work done by the car.

CAUTION ⚠ : I WANT A RIGHT ANSWER, IF YOU DON'T HAVE ANSWER DON'T WASTE MY TIME ⌚ ‍♀️.
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Answer 1

$\huge {\underline{\pink{Given:-}}}$

• Mass ,m = 3 × 10³ kg

• Initial velocity ,u = 54km/h = 54×5/18 = 15m/s

• Final Velocity ,v = 0m/s

• Distance travelled,s = 100m

$\huge {\underline{\green{To Find:-}}}$

• Work done by the Car ,W

$\huge {\underline{\blue{Solution:-}}}$

Firstly we calculate the acceleration of the car

Using 3rd Equation of Motion

• v² = u² +2as

Substitute the value we get

➨ 0² = 15² + 2×a × 100

➨ 0 = 225 + 200×a

➨ -225 = 200×a

➨ a = -225/200

➨ a = -1.125 m/s²

Here Negative sign show retardation

Therefore,the acceleration of the car is 1.125m/s².

As we know that Work Done is the Product of Force and displacement. and Force is the Product of mass and Acceleration.

• F = ma

Substitute the value we get

➨ F = 3×10³ × 1.125

➨ F = 3000 × 1.125

➨ F = 3375 N

And

• W = Fs

Substitute the value we get

➨ W = 3375 × 100

➨W = 337500

➨ W = 337500/1000 KJ

➨ W = 337.5 KJ

Therefore, the work done by the car the car is 337.5 KiloJoules.

Answer 2

• v² = u² +2as

Substitute the value we get

=> 0² = 15² + 2×a × 100

=> 0 = 225 + 200×a

=> -225 = 200×a

=> a = -225/200

=> a = -1.125 m/s²

• F = ma

Puttingthe value we get

=> F = 3×10³ × 1.125

=> F = 3000 × 1.125

=>> F = 3375 N

And

• W = Fs

Putting the value we get

=> W = 3375 × 100

=>W = 337500

=> W = 337500/1000 KJ

=> W = 337.5 KJ