A train of mass 600metric tonn is pulled by a locomotive of mass 150 metric ton . THe speed of the train is 54km /h . The locomotive pulls the train on level track . The force of friction on the locomotive and the train is 10N per metric ton . Calciulate the power of the locomotive
152.5W
100W
112.W
120W

A train of mass 400 tons climbs up an incline of 1//49 at the rate of 36kmh^(-1) . The force of friction is 4kg' at per ton. Find the power of the engine.

An engine is pulling a train of mass m on a level track at a uniform speed u . The resistive froce offered per unit mass is f .

A train rounds an unbanked circular bend of radius 30 m at a speed of 54 km//h . The mass of the train is 10^(6) kg. What provides the centripetal force required for this purpose - The engine or the rails ? What is the angle of banking required to prevent wearing out the rails ?

A train of mass 100 metric tons is ascending uniformly on an incline of 1 in 250, and the resistance due to friction, etc is equal to 6 kg per metric ton. If the engine be of 7.84xx10^(4) watts and be working at full power, find the speed at which the train is going

An engine of power 58.8 k W pulls a train of mass 2xx10^(5)kg with a velocity of 36 kmh^(-1) . The coefficient of kinetic friction is

Two trains A and B each of mass m are moving on a longitude of a spherical planet of mass M and radius R with velocity v relative to to planet. Train A & B presses thhe rail track with forces N_(A) & N_(B) respectively. [ omega is angular velocity of planet about its vertical axis] Then

Two block tied with a massless string of length 3m are placed on a rotating table as shown. The axis of rotation is 1m from 1kg mass and 2m from 2kg mass. The angular speed omega=4rad//s . Ground below 2kg block is smooth and below 1kg block is rough. (g=10m//s^(2)) (a) Find tension in the string, force of friction on 1kg block and its direction. (b) If coefficient of friction between 1kg block and groung is mu=0.8 . Find maximum angular speed so that neither of the blocksd slips. (c) If maximum tension in the string can be 100N , then maximum angular speed so that neither of the blocks slips.

Consider a wave sent down a string in the positive direction whose equation is given is y=y_(0)sin[omega(t-x/v)] The wave is propagated along because each string segment pulls upward and downward on the segment adjacent to it a slightly larger value of x and, as a result does work upon the string segment to which wave is travelling. For example, the portion of string at point A is going upward, and will pull the portion at point B upward as well. In fact, at any point along the string, each segment of the string is pulling on the segment just adjacent and to its right, causing the wagve to propagate. It is by this process that the energy is sent along the string. Now we try to calculate how much energy is propagated down the string per second T_(y)=tsintheta~~TtanthetaimpliesT_(y)=-T(dely)/(delx) (The negative sign appears because as shown in the figure II, the slope is negative) the force will act through a distance dy=v_(y)dt=(dely)/(delt) dt Therefore work done by force in time dt is dW=T_(y)dy=-T((dely)/(delx))((dely)/(delt))dtimpliesdW=(omega^(2)y_(0)^(2)T)/vcos^(2)[omega(t-x/V)]dt .....(A) How much average power is transmitted dawn a string having density mu=5xx10^(-4)kg//m and T=5N by a 200Hz vibration of amplitude of 0.20cm

A train of weight 10^(7)N is running on a travel track with uniform speed of 36 km h^(-1) . The frictional force is 0.5 kg f per quintal. If g = 10 ms^(-2) , power of engine is

A train of mass 2xx10^(5) kg has a constant speed of 20 ms^(-1) up a hill inclined at theta=sin^(-1)((1)/(50)) to the horizontal when the engine is working at 8xx10^(5) W. Find the resistance to motion of the trian. (Take ,g=9.8 ms^(-2) ).