Question 1182826
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A 2 kg block is pulled with uniform speed up a plane which makes 30 degrees with the horizontal by a force of 15 N 
which is parallel to the plane. Find the coefficient of kinetic friction between the block and the plane.
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<pre>
Since the block is pulled uniformly up the plane, it means that the force of 15 N parallel to the plane
is equal to the sum of the rolling force and the friction force.


The rolling force  R  parallel to the inclined plane is  R = m*g*sin(30°) = 2*10*(1/2) = 10 newtons.


The normal reaction force N, which is PERPENDICULAR to the inclined plane, has the value of  


    N = m*g*cos(30°) = {{{2*10*(sqrt(3)/2)}}} = {{{10*sqrt(3)}}} = 17.32 newtons.


The friction force  {{{F[fr]}}}  is equal to  {{{F[fr]}}} = k*N = 17.32*k,  where "k" is the kinematic friction coefficient.


So we have this equation to find out the kinetic friction coefficient


    15 = R + {{{F[fr]}}} = 10 + 17.32k.


From the equation, we find


    k = {{{(15-10)/17.32}}} = 0.287   (the dimensionless value).


<U>ANSWER</U>.  Under given condition, the kinetic friction coefficient value is  0.287.
</pre>

Solved.