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A crane is raising a load of 500 tonnes at a steady rate of 5 cm/s. what power is the engine of the crane producing?
Original post by monika16
A crane is raising a load of 500 tonnes at a steady rate of 5 cm/s. what power is the engine of the crane producing?


P=FvP=Fv

The unit of v is ms1ms^{-1} so you have a unit conversion to do. If the load weighs 500 tones, what force is being exerted against gravity?
Reply 2
Alternatively, the GPE gained in one second is (0.05×500000g)J (0.05\times 500000g) \text{J} . And power = work done/time.
This just gives a different way of thinking about it but in reality it is exactly the same as the method shown above.
(edited 7 years ago)
Reply 3
Original post by Kvothe the arcane
P=FvP=Fv

The unit of v is ms1ms^{-1} so you have a unit conversion to do. If the load weighs 500 tones, what force is being exerted against gravity?


To see why both are equivalent: P=Work donetime=Fst=Fst=FvP = \frac{\text{Work done}} { \text{time} } = \frac{Fs}{t} = F\frac{s}{t} = Fv.
(edited 7 years ago)

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