(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

1.a block of mass 5kg moves on a rough horizontal plane with coefficient of friction (Dynamic) 0.2 under the action of a horizontal force of 30N. If the block starts from rest , find the distance it travels in the first 3 second.

2.a particle of mass 6kg,moving at 8m/s on a smooth horizontal surface, goes onto a rough horizontal surface with a coefficient of friction 0.25> Find the distance it moves across the rough surface before coming to rest.

3.This is my own question.Because I might get the concept of friction force wrongly

It's been in my mind for about 3 days thinking of it

A particle moving up a slope with a maintained speed of 1m/s

In another situation a same particle moving up the same slope with a maintained speed of 5m/s

Do the two particle use the same force?

As the Coefficient,mass,gravity,angle of slope are the same

2. Relevant equations

F/N =u(Mew)

F=ma

s=ut+1/2 at^2

Anyhow Regarding the equation,

Relating The Momentum and Force

F=ma

F=m x v/t

Ft=mv

So If The Force is 15N and the object moving at a constant velocity last for 3 seconds

Ft=mv

15(3) = MV

It will be 45 kg m/s ??

3. The attempt at a solution

1.

M=5kg

Force=30N

μDynamic = 0.2

μStatic = F/N (Force Used To Overcome the Static Friction)/(Normal Force)

= 30/(30x9.8)

= 0.1020

Initial Velocity=0

Final Velocity = 0

Time=3 seconds

Distance = Unknown

Untill This Step

What Should I Do With the two different coefficient of friction??

2.

Upon This Stage I Do Realize that I'm Getting Stupiddd

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# Coefficient of Friction

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