Snapsolve any problem by taking a picture.
Try it in the Numerade app?

This problem has been solved

You'll get a detailed solution from a subject matter expert that helps you learn core concepts.

Question

A train weighing 1200 tons is pulled up a 2% grade at 10 mph. If the train resistance is constant at 12 lb per ton and the power exerted by the locomotive remains constant, what will be the speed of the train on a level track? What horsepower does the locomotive develop?

          
            A train weighing 1200 tons is pulled up a 2% grade at 10 mph. If
the train resistance is constant at 12 lb per ton and the power
exerted by the locomotive remains constant, what will be the speed
of the train on a level track? What horsepower does the locomotive
develop?

Added by Craig P.

Instant Answer

Step 1

Step 1: Calculate the weight of the train in kg: Given weight of the train = 1200 tons = 1200 * 1000 kg = 1,200,000 kg Show more…

Show all steps

Adi S and 81 other subject Physics 101 Mechanics educators are ready to help you.

Ask a new question

Recommended Videos

Recommended Textbooks

Transcript

00:02 In the given problem, we have the weight given to us in terms that is weight is equals to 1 ,200 tons.

00:11 So if we write the weight in cages in kilograms, we will say that the mass will be equals to 1 ,200 tons that will be multiplied to 1 ,000 kilograms.

00:24 Therefore, it will be 1204 times 0 kilogram.

00:31 This is the weight in kilograms.

00:34 The friction, according to the question, the friction that is the resistance applied is 12 lb per turn, that is pound per ton.

00:47 Therefore the friction, suppose it is in order as f s, will be equals to 12 pound per ton and in order to cancel this turn we can multiply it to 1 ,200 turn so that this turn and this turn can get cancelled and we will have the value of the friction to be 144 double zero as this is in pound so in new turn this will come out to be 6454 .39 newton this value of the friction is in newton the given velocity is 10 10 miles per hour, so in meter per second, the velocity will be 4 .4704 meter per second.

01:36 Therefore, the power required for 2 % grade track will be given as the power will be equal to v.

01:50 Sin theta multiplied to g multiplied to the mass.

01:54 That will be equals to v is 4 .4704 and the value of sine theta will be 0 .0 to 2 % will be 0 .02 and this will be the component that we have in the triangle that is 1 square plus 0 .02 whole square multiplied to the g that is 9 .81 and this will be multiplied to 120 and 4 zeros.

02:27 Thus the power over here will be 1 .05528 .87 volt.

02:36 This is the power over here...