question is kind badly put, i think
they mean driving force besides the weight
total work is zero, because velocity is constant
so WDF+Wweight=Wresistent
WDF+2700000=4800000
if i answered this i would consider the weight also a driving force
Q) A lorry of mass 15 000 kg moves with constant speed 14ms−1 from the top to the bottom of a straight hill of length 900 m. The top of the hill is 18m above the level of the bottom of the hill. The total work done by the resistive forces acting on the lorry, including the braking force, is 4.8 × 106 J.
Find
(i) the loss in gravitational potential energy of the lorry
(ii) the work done by the driving force.
My Attempt
The answer to part (i) is mgh = 2700000J
But I am stuck at (ii)
I used the formula
WDF - Work Done By Resistance = 2700000
WDF = 2700000+4.8x10^6
WDF = 7500000J
But the correct answer is 2.1 x 10^6J
How? Am I using a wrong method?
as the velocity is constant the
so the total work is zero
there are 3 forces
one is the weight which contributes for the truck to move forward and its work its equal to (as it contributes for the truck moving forward its work is positive) (Wweight= )
the second is the 'driving' force, it also contributes for the truck moving forward so its work is also positive (this is waht you want to know)
the third is the resistive forces which contribute to diminish the truck speed. so its work is negative (Wresistent=-4800000J)
as the total work is zero
WDF+Wweight+Wresistent=0