A train of mass 2-0- 105 kg has a constant speed of 20 m-s up a hill inclined at - - sin -1 1-50 to the horizontal when the engine is working at 8-0- 105 W- Find the resistance to motion of the train- g- 9-8 m-s2

The correct option is C 800 NSince, P=Fv F= P/v= 8.0× 10^5/20= 4.0× 10^4N At constant speed, the forces acting on the train are in equilibrium, Resol ...

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Standard XII

Physics

Law of Conservation of Mechanical Energy

A train of ma...

Question

A train of mass $2.0 \times 10^{5}$ kg has a constant speed of 20 m/s up a hill inclined at $θ = {sin}^{- 1} (\frac{1}{50})$ to the horizontal when the engine is working at $8.0 \times 10^{5}$ W. Find the resistance to motion of the train. $(g = 9.8 m / s^{2})$

400 N

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800 N

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1600 N

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1200 N

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Solution

The correct option is C 800 N
Since, $P = F v$
$F = \frac{P}{v} = \frac{8.0 \times 10^{5}}{20} = 4.0 \times 10^{4} N$
At constant speed, the forces acting on the train are in equilibrium, Resolving the forces parallel to the hill.
$F = R + (2.0 \times 10^{5}) g \times \frac{1}{50}$
$4.0 \times 10^{4} = R + 39200 o r R = 800 N$
Therefore, the resistance is 800 N.

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A train of mass 2.0×105 kg has a constant speed of 20 m/s up a hill inclined at θ=sin−1(150) to the horizontal when the engine is working at 8.0×105 W. Find the resistance to motion of the train. (g=9.8m/s2)

The correct option is C 800 NSince, P=FvF=Pv=8.0×10520=4.0×104NAt constant speed, the forces acting on the train are in equilibrium, Resolving the forces parallel to the hill.F=R+(2.0×105)g×1504.0×104=R+39200orR=800NTherefore, the resistance is 800 N.

A train of mass $2.0 \times 10^{5}$ kg has a constant speed of 20 m/s up a hill inclined at $θ = {sin}^{- 1} (\frac{1}{50})$ to the horizontal when the engine is working at $8.0 \times 10^{5}$ W. Find the resistance to motion of the train. $(g = 9.8 m / s^{2})$