Question

A train of mass 2.0 x 10^5 kg moves at a constant speed of 72 km h-' up a straight inclined plane
against a frictional force of 1.28 x 10^4 N. The inclination is such that the train rises vertically 1.0 m
for every 100 m travelled along the inclination. Calculate (i) the rate of increase per second of the
potential energy of the train, (ii) the necessary power developed by the train.​

Answer 1

train of mass

kg has a constant speed of 20

up a hill inclined at

to the horizontal when the engine is working at

W. Find the resistance to motion of the trian. (Take ,g=9.8

Answer 2

Concept:-

It might resemble a word or a number representation of the quantity's arithmetic value. It could resemble a word or a number that represents the numerical value of the quantity. It could have the appearance of a word or a number that denotes the quantity's numerical value.

Given:-

The given expression "A train of mass $2.0 \times 10^5$ kg moves at a constant speed of $72$ km h-' up a straight inclined plane against a frictional force of $1.28 \times 10^4$ N. The inclination is such that the train rises vertically $1.0$ m for every $100$ m travelled along the inclination."

Find:-

We need to find that Calculate $(i)$ the rate of increase per second of the potential energy of the train, $(ii)$ the necessary power developed by the train.​

Solution:-

Mass of the train $(m) = 2\times 10^5$ kg

Velocity of train $(v) =\frac{72\times 1000}{60\times 60}=20$m/s

Frictional force $= 1.28 \times 10^4$ N

Sin$\phi = ph=\frac{1}{100}$

The force against which work is to be done is given by

F $=$ $mgsin\phi +$ Frictional force

$= 2\times10^5\times 9.8\frac{1}{100}+ 1.28 \times 10^4\\= 3.28 \times 10^4$ N

$(i)$ the rate of increase per second of the potential energy of the train is $3.28\times 10^4$ N.

$(ii)$ Power developed in the train $= Fv = 3.28 \times 10^4 \times 20 = 6.56\times 10^5$ W.

Hence, the rate of increase per second of the potential energy of the train is $3.28\times 10^4$ N and Power developed in the train is $6.56 \times 10^5$.

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