Question

A car was moving with a constant velocity of 50 metre per second after brakes are applied travelled a distance of 30 metre then find the work done by breakes if mass of the car 1500 kg​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

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

• Initial velocity (u) = 50m/s.
• Final velocity (v) = 0 m/s.
• Distance travelled (s) = 30m.
• Mass of the Car (m) = 1500Kg.

$\huge{\bold{\underline{Explanation:-}}}$

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Firstly finding the Acceleration of the car,

Applying Third kinematic equation

$\large{\boxed{\tt v^2 - u^2 = 2as}}$

Substituting the values,

$\large{\tt \leadsto (0)^2 - (50)^2 = 2 \times a \times 30}$

$\large{\tt \leadsto 0 - 2500 = 60 \times a}$

$\large{\tt \leadsto - 2500 = 60a}$

$\large{\tt \leadsto a = \dfrac{-2500}{60}}$

$\large{\tt \leadsto a = \dfrac{\cancel{-2500}}{\cancel{60}}}$

Simplifying,

$\large{\boxed{\tt a = \dfrac{- 125}{3} \: m/s^2}}$

So, The acceleration is - 125/3 m/s².

Note:-

Here the Acceleration is negative as the body is retarding.

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Applying Newton's second law of motion,

$\large{\boxed{\tt F = ma}}$

Substituting the values,

$\large{\tt \leadsto F = 1500 \times \dfrac{-125}{3}}$

$\large{\tt \leadsto F = \cancel{1500} \times \dfrac{-125}{\cancel{3}}}$

$\large{\tt \leadsto F = 500 \times - 125}$

$\large{\boxed{\tt F = - 62,500 \: N}}$

So, the Force applied by the brakes is - 62,500N.

Note:-

Note:- Here the Force is negative as the body is retarding.

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Applying Work done formula,

$\large{\boxed{\tt W = F.s}}$

Substituting the values,

$\large{\tt \leadsto W = - 62,500 \times 30}$

$\large{\tt \leadsto W = - 18,75,000}$

Taking Magnitude

$\large{\tt \leadsto W = 18,75,000}$

$\huge{\boxed{\boxed{\tt W = 1.875 \times 10^6 \: J}}}$

So, the work done by the brakes is 1.875 × 10⁶ Joules.

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