A loaded lorry of total mass 5000 kg can come down from the top of a slope (1:40) effortlessly at `18 km *h^(-1)` . What should be the horsepower of its engine so that it can go up with the same speed, from the base to the top ? Resistance due to friction may be taken to be the same in both cases.

Explanation

Velocity of the lorry
`=18km*h^(-1) =(18xx1000)/(60xx60) m*s^(-1) =5 m*s^(-1)`
`sin theta =1/(40)`
As the speed of the lorry remains constant for the downward journey, the friction just balances the downward active force mg `sin theta`.
`therefore` Frictional force,`f=mg sin theta`
Effective force on the lorry for its upward journey
`= mg sin theta +f =mg sin theta +mg sin theta =2mg sin theta `
`therefore` power of the engine
= effective force on the lorry `xx` velocity of the lorry
`=2mg sin theta xx5 =2 xx5000xx9.8xx1/(40)xx5`
`=12250W (12250)/(746)hp =16.42 hp`.