It is given,
v=80kmh−1
m=500kg
S=4km
μ=0.04
The power due to friction is
Pf=Fv where F=μmg
=((0.04)(500×9.8)×80×185)W
Hence, the work done by the engine in overcoming the friction is
W=Pft=Pfvd
⇒W=20×9.8×80×185×80×1854×103=784kJ
JEE Main 2023 — Physics Mechanics
To maintain a speed of 80kmh−1 by a bus of mass 500kg on a plane rough road for 4km distance, the work done by the engine of the bus will be _____ kJ. [The coefficient of friction between tyre of bus and road is 0.04]
Held on 12 Apr 2023 · Verified 6 Jul 2026.
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