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A Level work done question

The actual efficiency of the car is 25%. The car takes 18 kg of petrol to fill its tank. The
energy provided per kilogram of petrol is 46 MJ kg–1. The drag force acting on the car at a
constant speed of 30 m s–1 is 500 N.
1 Calculate the work done against the drag force per second.
The mark scheme says to find the work done, it is 500 x 30. My question is why is it being multiplied by a speed when work done is equal to force x distance?
Original post by znx
The actual efficiency of the car is 25%. The car takes 18 kg of petrol to fill its tank. The
energy provided per kilogram of petrol is 46 MJ kg–1. The drag force acting on the car at a
constant speed of 30 m s–1 is 500 N.
1 Calculate the work done against the drag force per second.
The mark scheme says to find the work done, it is 500 x 30. My question is why is it being multiplied by a speed when work done is equal to force x distance?


Asking for work done against the force every second.
Speed = Distance/Time; Distance = Speed*Time
Work Done = Distance*Force = Speed*Time*Force = (30)*(1)*(500)
Reply 2
Original post by britishtf2
Asking for work done against the force every second.
Speed = Distance/Time; Distance = Speed*Time
Work Done = Distance*Force = Speed*Time*Force = (30)*(1)*(500)


Thank you very much that makes a lot of sense now :biggrin:
Reply 3
You are correct in saying that work done = force x distance, but this specific question is asking you how much energy the car is using 'per second' to overcome the resistive force. Therefore the correct thing to do is 500N x 30ms-1, because the speed 30ms-1 literally means '30 metres travelled per second'.
Original post by znx
Thank you very much that makes a lot of sense now :biggrin:


No problem. :smile:
There is a second question to this on my worksheet that says: calculate the total distance the car can travel on a full tank of petrol when travelling at a constant speed of 30m/scould someone please explain what i need to do here. many thanks

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