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June 14th, 2017, 12:46 PM  #1 
Newbie Joined: Jun 2017 From: Ferndown Posts: 2 Thanks: 0  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 Thanks Last edited by skipjack; June 15th, 2017 at 01:05 PM. 
June 14th, 2017, 05:34 PM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449  Quote:
net work = (net force)(displacement) = change in Kinetic Energy $(F_e  f  Mg\sin{\theta})\Delta x = \dfrac{1}{2}M(v_f^2v_0^2)$ $F_e \cdot \Delta x = \dfrac{1}{2}M(v_f^2v_0^2) + (f + Mg\sin{\theta}) \Delta x$ $M$ = 250 metric tons = 250000 kg $v_f = 45 \, km/hr = 12.5 \, m/s$ $g = 9.8 \, m/s^2$ $\sin{\theta} = \dfrac{1}{\sqrt{10001}}$ $F_e \cdot \Delta x = \dfrac{1}{2}(250k)(12.5^20) + (30k + 250k \cdot g\sin{\theta})(250)$ $F_e \cdot \Delta x \approx 3.32 \times 10^7 \, J$  
June 15th, 2017, 10:18 AM  #3 
Newbie Joined: Jun 2017 From: Ferndown Posts: 2 Thanks: 0 
ah I see! Thank you very much for your help!


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conservation, energy, inclined, inclined plane, plane, problem, work, work done 
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