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Questions  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

A box can be moved up an inclined plane with constant velocity by a force of magnitude $F_1$ or down the inclined plane with constant velocity by a force of magnitude $F_2$ . Find the coefficient of kinetic friction $\mu_k$ between the box and the inclined plane. It holds that $F_1=6F_2$ and both forces are parallel with the inclined plane. The angle $\alpha$ between the inclined plane and the horizontal plane is $15^{\circ}$

$\begin{array}{1 1}0.48\\0.36\\0.24 \\0.62\end{array} $

1 Answer

Step 1:
Draw the free body diagrams :
The box is being moved up :
The box is being moved down :
Step 2 :
The force $F_G$ decomposes into two components in the direction of these axes :
$F_{Gx} =F_G \sin \alpha$
$F_{Gy} =F_G \sin \alpha$
The box is moved up :
$x: -F_G \sin \alpha +F_1- F_t=0$ ----(1)
$y : N- F_G \cos \alpha =0$-----(2)
The box is moved down :
$x: -F_G \sin \alpha +F_2- F_t=0$ ----(3)
$y : N - F_G \cos \alpha =0$-----(2)
Step 3:
The friction force that affects the box is directly proportional to the normal force by which the box affects the inclined plane .
This force is , according to Newton's third law, of the same magnitude as the force by which the inclined plane affects the box .
The friction force$F_t$ can therefore be expressed as follows :
$F_t= \mu _k N$ ------(4)
We express the force $F_t$ can therefore be expressed as follows :
We express the force N from equation (2) :
$N= F_G \cos \alpha$
Equation for the friction force :
$F_t = \mu _k F_G \cos \alpha$ -------(5)
Step 4 :
Determining the coefficient of kinetic friction $\mu_K$
We substitute into motion equation (1) and (3) the equation for the friction force (5).
We express the force $F_1 $ as $6 F_2$
$-F_G \sin \alpha +6 F_2- F_G \mu _K \cos \alpha =0$ -------(6)
$F_G \sin \alpha + F_2 -F_G \mu _K \cos \alpha =0$ -------(7)
Equation (7) is multiplied by 6 and subtracted from equation (6):
$7 F_G \sin \alpha - 5 F_G \mu _k \cos \alpha =0$
$7F_G \sin \alpha \alpha = 5 F_G \mu _K \cos \alpha /: (5 \;F_G \cos \alpha)$
$\mu_K =\large \frac{7}{5} $$ \tan \alpha$---(8)
For the given numerical value we get from (8):
$\mu_k =\large\frac{7}{5} $$ \tan 15^{\circ}$
$\qquad= \large\frac{7}{5}$$. 0.258$
$\qquad= 0.36 $
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