Mechanics - work done conservation of energy *Help needed*

Click For Summary
A train with a mass of 250 tonnes accelerates up a 1 in 100 incline, reaching 45 Kph after 200m, while facing a constant frictional resistance of 30KN. The discussion focuses on calculating the work done by the engine using conservation of energy principles rather than force and kinematic equations. Participants emphasize the importance of accounting for both potential energy changes due to elevation and kinetic energy changes from speed. The correct approach involves equating the work done by the engine and friction to the total energy change in the system. Clarification is sought on how to properly apply these energy equations to solve the problem.
EddyH
Messages
2
Reaction score
0

Homework Statement


A train with a mass of 250 tonnes starts from rest and accelerates up an incline of 1 in 100 attaining a speed of 45 Kph after traveling 200m. If the frictional resistance to motion is constant at 30KN calculate the work done by the engine using the principle of conservation of energy

Homework Equations


Possible relevant equations:

Work done= Force*Distance
Force=Mass*acceleration
Kinetic energy = 1/2*Mass*Velocity^2
Potential energy= Mass*Gravity (9.81)* Height

The Attempt at a Solution



Force= 250*103 Kg*12.5= 3 125 000 kg/m/s
Work done= 3 125 000*200= 625 000 000(J)
Not sure whether this is along the right lines or not. I am also unsure where the frictional resistance and gravity is used.

Thank you for your time.
 
Physics news on Phys.org
EddyH said:

Homework Statement


A train with a mass of 250 tonnes starts from rest and accelerates up an incline of 1 in 100 attaining a speed of 45 Kph after traveling 200m. If the frictional resistance to motion is constant at 30KN calculate the work done by the engine using the principle of conservation of energy

Homework Equations


Possible relevant equations:

Work done= Force*Distance
Force=Mass*acceleration
Kinetic energy = 1/2*Mass*Velocity^2
Potential energy= Mass*Gravity (9.81)* Height

The Attempt at a Solution



Force= 250*103 Kg*12.5= 3 125 000 kg/m/s
net force is mass times acceleration. You are not using the net force or the acceleration.
Work done= 3 125 000*200= 625 000 000(J)
Your incorrect values are calculating the net total work done. You are looking for the work done by the engine only.
Not sure whether this is along the right lines or not. I am also unsure where the frictional resistance and gravity is used.

Thank you for your time.
the problem is asking you to use conservation of energy, not force and kinematic equations. What is the conservation of energy equation that relates work and energy?
 
Sorry to be a pain, but I do not know how to go about this, please can you explain?
Thank you
 
Say, Eddy, if you are asked to solve the problem using energy methods, you should know about the possible energy equations to use, for example, you should know that the work done by non conservative forces (like the engine force and friction force in this example) must equal the change in PE plus the change in KE of the system. Give it a try.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
7
Views
2K
Replies
15
Views
2K
Replies
9
Views
934
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
20
Views
3K