How to Calculate Distance on an Incline with Friction

[f(x)]

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 ?

 

[Chi Meson]

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?

 

[f(x)]

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.

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.

 

[DAKONG]

i think P+mgsinx-umgcosx=ma so..
it's nothing ; enhance force

 

[Doc Al]

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 ?

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.

 

[f(x)]

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

 

[Weimin]

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.

 

[Doc Al]

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.

 

[f(x)]

Ahh.. fine. Thx for the explanation