To maintain a speed of 80 km/h by a bus of mass 500 kg on a plane rough road for 4 km distance, the work done by the engine of the bus will be KJ. [The coefficient of friction between tyre of bus and…

To maintain a speed of 80 km/h by a bus of mass 500 kg on a plane… — Solution | JEE Physics

NVAEasyJEE 2023Work Done by Force

To maintain a speed of $80 \, \text{km/h}$ by a bus of mass $500 \, \text{kg}$ on a plane rough road for $4 \, \text{km}$ distance, the work done by the engine of the bus will be _____ $\text{KJ}$ . [The coefficient of friction between tyre of bus and road is $0.04$ ].

Answer

Correct answer:784

Step-by-step solution

Standard Method

Given: Speed is constant at $80 \, \text{km/h}$ , mass of bus is $500 \, \text{kg}$ , coefficient of friction is $\mu = 0.04$ , and distance is $x = 4 \, \text{km} = 4 \times 10^3 \, \text{m}$ .

Find: The work done by the engine.

For constant speed, the change in kinetic energy is zero. Hence, by the work-energy theorem, work done by the engine and work done by friction must cancel each other.

\text{WD}_{\text{engine}} + \text{WD}_{\text{friction}} = 0

Therefore,

\text{WD}_{\text{engine}} = -\text{WD}_{\text{friction}} = -\left[\mu m g x\right]

Using $\mu = 0.04$ , $m = 500 \, \text{kg}$ , $g = 9.8 \, \text{m/s}^2$ , and $x = 4 \times 10^3 \, \text{m}$ ,

\text{WD}_{\text{engine}} = -0.04 \times 500 \times 9.8 \times 4 \times 10^3 = 784 \, \text{KJ}

Thus, the work done by the engine is $784 \, \text{KJ}$ . Therefore, the required numerical answer is 784.

Common mistakes

Using the given speed $80 \, \text{km/h}$ in the calculation. This is wrong because for constant speed on a rough horizontal road, the work done here depends on friction force and distance, not on the numerical value of speed. Use $W = F \cdot x$ with friction force.
Forgetting to convert $4 \, \text{km}$ into metres. This is wrong because friction force is in SI units, so distance must be in metres. Use $x = 4 \times 10^3 \, \text{m}$ .
Taking friction work as positive for the road on the bus. This is wrong because friction opposes motion, so its work on the bus is negative. Then the engine must do equal positive work to maintain constant speed.

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