How to Calculate Distance on an Incline with Friction

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Homework Help Overview

The problem involves calculating the distance a block of mass 4 kg moves down an incline under the influence of a pushing force and friction. The incline has a coefficient of kinetic friction of 0.11 and is set at an angle of 30 degrees. The block starts from rest and the scenario is analyzed over the first 2 seconds of motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the role of the pushing force and its impact on the equations of motion, questioning whether it should be considered as acting continuously or only instantaneously. There is also exploration of how to combine the forces acting on the block to determine the net force.

Discussion Status

The discussion is ongoing with participants offering different perspectives on the nature of the pushing force. Some suggest that the push should be treated as continuous, while others explore the implications of it being instantaneous. Guidance has been provided regarding the need to consider impulse when discussing instantaneous forces.

Contextual Notes

There is some confusion regarding the treatment of the pushing force, with participants questioning the assumptions made about its duration and effect on the block's motion. The problem context does not clarify whether the push is sustained or momentary, leading to varied interpretations.

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Homework Statement


A block of mass = 4kg is pushed down an incline with a force of 4 N. The coefficient of kinetic friction of the incline = 0.11 . How far will the block move on the incline in the first 2 seconds after starting from rest ?
Given that angle of incline = 30 degrees.


Homework Equations


mgsinx-umgcosx=ma

The Attempt at a Solution



I am having trouble understanding the role of the PUSH. since the force due to the push acts only for a moment, how do I account for it in the equations of motion ?
 
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mgsinx represents one of the forces parallel to the surface of the plane.

umgcosx represents another (do you know which one?)

THere is a third force (the pushing force). How does this third force join the first two to give you the net force?
 
Chi Meson said:
mgsinx represents one of the forces parallel to the surface of the plane.

umgcosx represents another (do you know which one?)
umgcosx is the force of friction.

Chi Meson said:
THere is a third force (the pushing force). How does this third force join the first two to give you the net force?
The pushing force is instantaneous...so it will be there in my eqn. at one instant but not after that ? I feel its P+mgsinx-umgcosx=ma , where P=pushing force.
 
i think P+mgsinx-umgcosx=ma so..
it's nothing ; enhance force:rolleyes:
 
f(x) said:
I am having trouble understanding the role of the PUSH. since the force due to the push acts only for a moment, how do I account for it in the equations of motion ?

f(x) said:
The pushing force is instantaneous...

Somehow you got it into your head that the push acts just for an instant, but that's NOT what the problem says. Assume the push is maintained as the block goes down the incline.
 
Doc Al said:
Somehow you got it into your head that the push acts just for an instant, but that's NOT what the problem says. Assume the push is maintained as the block goes down the incline.

Yeah this gives me the correct answer which is 10 m.
But i have a query,how would you solve this if the force was only instantaneous ?

Thx for the help ChiMeson and DocAl
 
In the case of "instantaneous", you still need to know the time the force acts on the block in order to calculate how much momentum the block gains from the force.
 
f(x) said:
But i have a query,how would you solve this if the force was only instantaneous ?
As Weimin says, you need to think in terms of the impulse that the force exerts, which produces a change in momentum of the block. The impulse = force*time that the force acts. (Even if it seems "instantaneous", to have an effect the force must act for some nonzero amount of time.) Once you have the change in momentum, you can calculate the speed of the block after the impact. Then it's just another force/acceleration problem, only now the block has some speed instead of starting from rest.
 
Ahh.. fine. Thx for the explanation :smile:
 

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