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4) \ \

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Train total mass m =1.8× 106 kg \ \

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Rising height h = 700 m \ \

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Traveling distance d = 53 km \ \

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d = 53000 m \ \

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Average speed v = 18 km/h \ \

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v = 18*(5/18) m/s \ \

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v = 5 m/s \ \

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The acceleration of gravity g = 9.81 m/s² \ \

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The frictional force is 0.8 percent of the weight \ \

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Ff = ( 0.8 / 100 ) mg \ \

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Ff = ( 0.8 / 100 )*1.8*106* 9.8 \ \

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Ff = 141120 \ \

\ The power output of the train Pt = ? \ \ \

The kinetic energy of the train KE \ \

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KE = ½mv² \ \

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KE = ½(1.8× 106)(5)²

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KE = 22.5× 106  J \ \

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At starting height is h1 = 0 m

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At ending height is h2 = 700 m

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Change in height  Δh = h2 - h1

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Δh = 700 - 0

\ Δh = 700 m \

The total change in potential energy ΔPE = mgΔh \ \

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ΔPE = (1.8× 106)(9.8)(700) \ \

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ΔPE = 12348 × 106  J \ \

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The energy dissipated due to friction Ef = Frictional force x distance \ \

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Ef = Ff x d \ \

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Ef = 141120 x 53000 \ \

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Ef = 7479360000 \ \

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Ef = 7479.36 × 106  J \ \

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Total energy Et = KE+PE+Ef \ \

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Substitute KE = 22.5× 106 , PE = 12348 × 106 , Ef = 7479.36 × 106 \ \

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Et = (22.5× 106)+(12348 × 106)+(7479.36× 106) \ \

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Et = 19849.86 × 106  J \ \

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Total force = Total energy / distance \ \

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Ft = Et / d \ \

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Ft = (19849.86 × 106 ) / 53000 \ \

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Ft = 0.37452566 × 106 N \ \

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Total power Pt = Total Force × velocity \ \

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Pt = (Ft)v \ \

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Pt = (0.37452566 × 106)(5) \ \

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Pt = 1.87263 × 106 W \ \

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Pt = 1.87263 MW \ \

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The power output of the train is 1.87263 MW \ \