Work Energy Power

JEE Main level, numerical answer, medium

150 of 183

Question 150 of 183

To maintain a speed of 80kmh−180km{h}^{-1} by a bus of mass 500kg500kg on a plane rough road for 4km4km distance, the work done by the engine of the bus will be _____ kJkJ. [The coefficient of friction between tyre of bus and road is 0.040.04]

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Answer: 784

It is given,
v=80kmh−1v=80km{h}^{-1}
m=500kgm=500kg
S=4kmS=4km
μ=0.04\mu =0.04
The power due to friction is
Pf=Fv{P}_{f}=Fv where F=μmgF=\mu mg
=((0.04)(500×9.8)×80×518)W=\left(\left(0.04\right)\left(500\times 9.8\right)\times 80\times \dfrac{5}{18}\right)W
Hence, the work done by the engine in overcoming the friction is
W=Pft=PfdvW={P}_{f}t={P}_{f}\dfrac{d}{v}
⇒W=20×9.8×80×518×4×10380×518=784kJ\Rightarrow W=20\times 9.8\times 80\times \dfrac{5}{18}\times \dfrac{4\times {10}^{3}}{80\times \dfrac{5}{18}}=784kJ

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