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Question
A truck weighing 1000 kgf changes its speed from 36 km h-1 to 72 km h-1 in 2 minutes. Calculate:
- the work done by the engine and
- its power (g = 10 m s-2)
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Solution
(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
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