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    A car exerts a driving force of 500N when travelling at a constant speed of 72 KMH-1 on a level track. What is the work done in 5 minutes??

    According to the marksceme the answer is 2.0x10^5

    Can someone explain how to get that that answer. Thanks !
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    I don't get that answer, what I get is :-

    Work done = Force * Distance.

    Distance travelled = speed * time = 72 * (5/60) = 6km = 6000m

    So, Work done - 500N * 6000m = 3x10^6 Nm

    Is that what you've been getting?
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    72 km h^-1 = 20 m s^-1

    in 300 secs the car travels 6000 m

    over that distance the car is exerting constant 500N force

    work done = Fs

    500 x 6000 = 3x10^6 J
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    72KmH-1 = 20m/s

    P=FV
    20*500= 10,000W
    PT=E
    (5mins= 300 secs)
    10,000*300= 3x10^6J
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    fwiw MS says 3.0 x 10^6 J too

    Q23 in paper http://filestore.aqa.org.uk/resource...081-SQP.PDF#24

    answer to 23 is A (3.0 x 10^6 J)
    in MS http://filestore.aqa.org.uk/resource...081-SMS.PDF#13
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    I think you've been reading the wrong bit of the mark scheme Kraixo - the answer is A - 3x10^6J
 
 
 
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