Solving Energy & Power Homework on Train Mass & Vertical Rise

[mizzy]

Homework Statement

A train of mass 2.4x10^6kg enters a 5km stretch of track with a vertical rise of 40m at a s speed of 1.0m/s and leaves at 3.0m/s. Assuming that frictional drag is negligible, find:

a) Each term of the equation Win = delta K + delta U + Wout

Homework Equations

Delta K = 1/2mvf^2 - 1/2mvi^2

Delta U = mghf - mghi

Wout = work removed from the system such as air drag.

Win = not sure??

The Attempt at a Solution

I found Wout and that's equal to 0.

I found Delta K. Not too sure if I used the right velocities. Final velocity, i used 3.0m's and inital velocity, i used 1.0m/s.

I found Delta U, with the final height being 40m and the initial height being 0m.

Am i doing these correctly? Can someone guide me please?

 

[Doc Al]

I found Wout and that's equal to 0.

I found Delta K. Not too sure if I used the right velocities. Final velocity, i used 3.0m's and inital velocity, i used 1.0m/s.

I found Delta U, with the final height being 40m and the initial height being 0m.

Looks like you're doing fine.

Since the energy increases, there must be some work input. That's what your equation will tell you.

 

[mizzy]

so is work input = delta k + delta U?

 

[Doc Al]

so is work input = delta k + delta U?

Yes.

 

[mizzy]

part b of the question asks for the average mechanical power delivered by the engine for the climb if it is done at constant acceleration.

I know power is work over time. We don't have time, but can we find that since they said it's done at constant acceleration? but don't know how to start...

 

[Doc Al]

What's the average speed during the climb?

 

[mizzy]

What's the average speed during the climb?

speed = d/t
where, speed is 1.0m/s and distance is 5m

so t = d/speed?

 

[Doc Al]

If the speed uniformly goes from 1 m/s to 3 m/s, what's the average speed?

 

[mizzy]

If the speed uniformly goes from 1 m/s to 3 m/s, what's the average speed?

Average speed is equal to the total distance traveled divided by time.

The average speed is 2km?

 

[Doc Al]

For uniformly accelerated motion, the average velocity = (v_i + v_f)/2.

See: https://www.physicsforums.com/showpost.php?p=905663&postcount=2"

 

[mizzy]

So to answer part b,

Work = delta K + delta U
= 1/2m(vf-v1)^2 + mg(h2-h1)

doing the calculation for Work, I got 9x10^8J.

Average speed is equal (vi + vf)/2,
therefore, it is 2m/s.

speed = distance/time
time = distance/speed
= 5/2
= 2.5s

so POWER = work / time

= 9x10^8 / 2.5
= 4x10^8Watts

Is that correct?

 

[Doc Al]

So to answer part b,

Work = delta K + delta U
= 1/2m(vf-v1)^2 + mg(h2-h1)

doing the calculation for Work, I got 9x10^8J.

I get closer to 9.5 x 10^8 J.

Average speed is equal (vi + vf)/2,
therefore, it is 2m/s.

OK.

speed = distance/time
time = distance/speed
= 5/2
= 2.5s

Careful. The distance is 5 km, not 5 m.

 

[mizzy]

I get closer to 9.5 x 10^8 J.


OK.


Careful. The distance is 5 km, not 5 m.

ok. That's a silly mistake. I should've noticed that. Thanks.

ok...here's what I get:

average speed = 2m/s
distance = 5000m

therefore, time = 2500s

POWER = 9.5x10^8J / 2500s = 3.8 x 10^5W

Thanks for your help.