For constant speed, work done by the engine \( \text{WD}_{\text{engine}} \) + work done by friction \( \text{WD}_{\text{friction}} = 0 \) (by Work-Energy Theorem).
Thus, we can write:
\[
\text{WD}_{\text{engine}} = -\text{WD}_{\text{friction}} = -\left[ \mu m g x \right]
\]
where \( \mu = 0.04 \), \( m = 500 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( x = 4 \, \text{km} = 4 \times 10^3 \, \text{m} \).
So,
\[
\text{WD}_{\text{engine}} = -0.04 \times 500 \times 9.8 \times 4 \times 10^3 = 784 \, \text{KJ}.
\]
Thus, the work done by the engine is 784 KJ.
A wheel of a bullock cart is rolling on a level road, as shown in the figure below. If its linear speed is v in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel, respectively) ?
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: