An engine pulls a $1500$ kg car on a level road at a constant speed of $5.0$ m $s^{- 1}$ against a frictional force of $500$ N. Calculate the power expended by the engine. What extra power has the engine to expend in order to maintain the same speed of the car up an inclined plane having a gradient of $1$ in $10$ ?

Open in App

Solution

Please refer work, power, energy related numericals
for this questions
wt.of car $= 1500 K g$
constant speed of car $= 5.0 m s^{- 1}$
then total force by the car to move for work =total force by engine + frictional force
$= 1500 \times 5 + 500$
$f_{1} = 7500 + 500$
$f_{1} = 8000 N \approx f_{1} = 8 K N$
Total work done $= w_{1} = F_{1} \times v = 8000 \times 5$
$W_{1} = 40, 000 w a t t$
$w_{1} = 40 k w$

case II
$\Rightarrow$ gradient $1$ in $10$ means
then $sin θ = \frac{1}{10}$ .
Total Applied force = $M g sin θ + F r i c t i o n a l f o r c e$
$F_{2} = 1500 \times 9.8 \times \frac{1}{10} + 500$ $\begin{matrix} A B = 10 B C = 1 \end{matrix}$
$F_{2} = 1970 N$

total work done $w_{2} = 1970 \times 5 = 9850 w a l r$
extra or $9.850 K w$
Total work done $w = w_{1} + w_{2}$
$= 40 + 9.85 = 49.85$

Suggest Corrections

Similar questions

An engine pull a 1500kg of car on a level road at a constant speed of 5m/s against a frictional force of 500N. Calculate the power expended by the engine. What extra power has the engine to expend in order to maintain the same speed of the car up an inclined plane having a gradient of 1in 10?

Q. An engine can pull