Kinetic energy ignoring friction

Click For Summary

Homework Help Overview

The problem involves two trucks with different masses being pushed with equal force for a set duration, specifically examining the resulting kinetic energy while ignoring friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between mass, force, acceleration, and kinetic energy. One participant attempts to analyze the accelerations and final velocities of the trucks based on Newton's second law.

Discussion Status

There is an ongoing exploration of the implications of mass on acceleration and kinetic energy. Some participants provide calculations and reasoning regarding the final velocities and kinetic energy of each truck, while others seek clarification on the concepts involved.

Contextual Notes

Assumptions include equal force applied to both trucks and the absence of friction. There is a mention of potential missing information that could affect the conclusions drawn.

David Earnsure
Messages
8
Reaction score
0

Homework Statement


2 trucks, one with twice as much mass than the other, are at rest on separate tracks. A man pushes each truck for 5 seconds. If you ignore friction and assume equal force is exerted on both trucks - which truck will have the most kinetic energy afterwards?

Homework Equations


KE=1/2mv^2

The Attempt at a Solution


Straight away I would say that the lighter truck would have more kinetic energy as it would take less force to move it, but as friction is being ignored and there is an equal force on both of them this stumps me, one of the options is that there is not enough information to answer the question but I feel like because friction is being ignored and there's equal force there is a really simple answer to this. I would appreciate someone steering me in the right direction so I can find the answer out for myself.

Thanks!
 
Physics news on Phys.org
How do the accelerations compare? The final velocities after 5 seconds?
 
Thankyou for your prompt reply,

Well since a = F/m and I know that the force is the same but that the masses are different, I can say that the acceleration of the heavy truck (x) would be half that of the lighter truck (y).
For example:
F=5
mass(x) = 2
mass(y) = 1

a(x) = 5/2 = 2.5
a(y) = 5/1 = 5

For the final velocity as both are stationary to begin with I can say that vf = at, meaning vf(x) would be half that of vf(y).

Since KE = 1/2m*v^2 if velocity is doubled KE will then be quadrupled (as velocity is squared) however as the lighter truck (y) is half the mass of heavier truck (x) it means that the KE of (y) is only double that of (x).

Therefore the lighter truck has more KE.

Thanks! (if I have somehow managed to go completely wrong with my explanation please let me know)
 
You can apply Newton's law too, F=dp/dt
 
David Earnsure said:
Thankyou for your prompt reply,

Well since a = F/m and I know that the force is the same but that the masses are different, I can say that the acceleration of the heavy truck (x) would be half that of the lighter truck (y).
For example:
F=5
mass(x) = 2
mass(y) = 1

a(x) = 5/2 = 2.5
a(y) = 5/1 = 5

For the final velocity as both are stationary to begin with I can say that vf = at, meaning vf(x) would be half that of vf(y).

Since KE = 1/2m*v^2 if velocity is doubled KE will then be quadrupled (as velocity is squared) however as the lighter truck (y) is half the mass of heavier truck (x) it means that the KE of (y) is only double that of (x).

Therefore the lighter truck has more KE.

Thanks! (if I have somehow managed to go completely wrong with my explanation please let me know)
Perfect!
 

Similar threads

Replies
24
Views
3K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
3K
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
18
Views
3K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K