Question

A truck weighing 1000 kgf changes its speed from 36 km h^-1 to 72 km h^-1 in 2 minutes. Calculate:

1. the work done by the engine and
2. its power (g = 10 m s^-2)

Answer 1

(i) Weight of truck = Force = 1000 kgf

∴ Mass of truck = `(1000 " kg" xx "g")/"g" = 1000 " kg"`

Initial Speed u = 36 km h^-1

= `36 × 5/18`

= 10 ms^-1

Final speed v = 72 km h^-1

= `72 xx 5/18`

= 20 ms^-1

Work done = Increase in energy

= `1/2 mv^2 - 1/2 m u^2`

= `1/2 m [(v + u) (v - u)]`

= `1/2 xx 1000 (20 + 10) (20 - 10)`

= `1/2 xx 1000 xx 30 xx 10`

Work done W = 150000 J = 1.5 × 10⁵ J

t = 2 minutes = 2 × 60 = 120s

(ii) Power of Engine = `w/t`

= `150000/120`

= 1250 W

= `1250/100 xx 100`

= `125/100 xx 10^3`

= 1.25 × 10³ W