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A truck of mass $1800 kg$ is moving with a speed $54 km / h$ . When brakes are applied, it stops with uniform negative acceleration at a distance of $200 m$ . Calculate the force applied by the brakes of the truck and the work done before stopping.

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200 kJ

202.5 kJ

151.75 kJ

101.25 kJ

answer is B.

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Detailed Solution

A truck of mass

1800 kg

is moving with a speed

54 km / h

. When brakes are applied, it stops with uniform negative acceleration at a distance of

200 m

. The force applied by the brakes of the truck and the work done before stopping is

202.5 kJ

.
Given:
Mass,

m = 1800 kg

Initial velocity,

u = 54 km / h

\Rightarrow

= 54 \times \frac{5}{18} m / s

\Rightarrow

= 15 m / s

Final velocity,

v = 0

Distance,

s = 200 m

from the 3rd equation of motion

v^{2} - u^{2} = 2 as

Retardation,

a = \frac{v^{2} - u^{2}}{2 s}

\Rightarrow

= \frac{0 - (15)^{2}}{2 \times 200}

a = \frac{- 9}{16} m / s^{2}

Force,

F = ma

\Rightarrow

= 1800 \times (\frac{- 9}{16})

F = - 1012.5 N

The negative sign indicates force acts in the opposite direction to motion.
Work done

= Fs

\Rightarrow

= - 1012.5 \times - 200

\Rightarrow = 202500 J

\Rightarrow

= 202.5 kJ

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