Question

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?

Answer 1

P1=Force×VelocityP_1 = Force × VelocityP1​=Force×Velocity
= 500 N × 5 m/s
= 2500 Watt
= 2.5 kilowatt

Power expended by engine is 2.5 kW

now,

$P_2=F\times V$

$=\frac{\left(1500\left(9.8\right)\left(5\right)\right)}{10}$

$=7350$

$=7.35kW$

total power=P1+P2total~power= P_1+P_2

Ptotal=9.85kWP_{total}=9.85 ktotal​

$P_{total}=P_1+P_2$

$=2.5+7.35$

$=9.85kW$