QJEEM23Ph1a80: JEE Main 2023 Physics - Class 11 JEEM23Ph1a80 Solution: To maintain a speed of 80 km/h by a bus of mass 500 kg on a…

To maintain a constant speed, the bus has to overcome the frictional force acting on it. The frictional force is given by:

$F_{friction} = \mu F_N$

Where $\mu$ is the coefficient of friction and $F_N$ is the normal force acting on the bus. Since the bus is on a flat road, the normal force is equal to the gravitational force:

$F_N = mg$

Where $m$ is the mass of the bus and $g$ is the acceleration due to gravity (approximately $9.8 \mathrm{~m/s^2}$ ).

Substituting the values, we get:

$F_{friction} = 0.04 \times 500 \times 9.8$

$F_{friction} = 196 \mathrm{~N}$

To maintain a constant speed, the engine must exert a force equal in magnitude to the frictional force. The work done by the engine to overcome the frictional force is given by:

$W = F_{friction} \times d$

Where $d$ is the distance traveled. First, convert the distance from km to m:

$d = 4 \mathrm{~km} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} = 4000 \mathrm{~m}$

Now, calculate the work done:

$W = 196 \mathrm{~N} \times 4000 \mathrm{~m}$ $W = 784000 \mathrm{~J}$

Convert the work done from joules to kilojoules:

$W = \frac{784000 \mathrm{~J}}{1000 \mathrm{~J/ kJ}} = 784 \mathrm{~kJ}$

The work done by the engine of the bus to maintain a speed of 80 km/h for a 4 km distance is $784.8 \mathrm{~kJ}$ .