[SOLVED] Work, Power and Energy - Mechanics

Sep 2009
148
2
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?
 
Last edited:
Jul 2009
68
20
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
 
Sep 2009
148
2
Huh? This is a past paper question so I doubt the examiner would make a mistake.

I am still not sure what you did or what to do with this sort of question!!
 
Jul 2009
68
20
\(\displaystyle W_{total}=\Delta E_c\)
as the velocity is constant the \(\displaystyle \Delta E_c=0\)

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 \(\displaystyle -\Delta E_p\) (as it contributes for the truck moving forward its work is positive) (Wweight=\(\displaystyle -\Delta E_p=2700000J\))
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
\(\displaystyle W_{total}=\)WDF+Wweight+Wresistent=0
 
Last edited:
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