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June 14th, 2017, 11:46 AM   #1
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Post Inclined plane problem (work done)

Hi guys, I have a question here that I am struggling with and it goes as follow:

A train of mass 250 tonnes starts from rest and accelerates up an incline of 1/100 attaining a speed of 45 km/h after travelling 250m. If the frictional resistance to motion is constant at 30kN calculate the work done by the train's engine using the principle of conservation of energy.

I have converted the speed to be be 12.5 m/s and tonnes into kg by x10^3. I think that the height is 2.5m as for every 100m it increases 1m and the total distance is 250m so 250/100 = 2.5m.

With this I calculated work done against gravity to be 6125000j (250x10^3 x 9.8 x 2.5).

Frictional work done I calculated as 30,000 x 250 to get 7500000N.

But I am now stuck from where to go on from here :/ I may just be being stupid, but any help would be greatly appreciated

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Last edited by skipjack; June 15th, 2017 at 12:05 PM.
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June 14th, 2017, 04:34 PM   #2
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Quote:
A train of mass 250 tonnes starts from rest and accelerates up an incline of 1/100 attaining a speed of 45 km/h after travelling 250m. If the frictional resistance to motion is constant at 30kN calculate the work done by the train's engine using the principle of conservation of energy.
work done by the train engine is $F_e \cdot \Delta x$ ...

net work = (net force)(displacement) = change in Kinetic Energy

$(F_e - f - Mg\sin{\theta})\Delta x = \dfrac{1}{2}M(v_f^2-v_0^2)$

$F_e \cdot \Delta x = \dfrac{1}{2}M(v_f^2-v_0^2) + (f + Mg\sin{\theta})
\Delta x$

$M$ = 250 metric tons = 250000 kg

$v_f = 45 \, km/hr = 12.5 \, m/s$

$g = 9.8 \, m/s^2$

$\sin{\theta} = \dfrac{1}{\sqrt{10001}}$

$F_e \cdot \Delta x = \dfrac{1}{2}(250k)(12.5^2-0) + (30k + 250k \cdot g\sin{\theta})(250)$

$F_e \cdot \Delta x \approx 3.32 \times 10^7 \, J$
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June 15th, 2017, 09:18 AM   #3
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ah I see! Thank you very much for your help!
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