8 questions · timed · auto-graded
Body is kept on the horizontal table.
If no force is applied, no frictional force will be there
f → frictional force
F → Applied force
From grap it can be seen that when applied force is zero,
frictional force is zero.
Let m = mass of the block
From the freebody diagram,
$\text{R}-\text{mg}=0\Rightarrow\text{mf} \ ...(1)$
Again $\text{ma}-\mu\text{R}=0\Rightarrow\text{ma}=\mu\text{R}=\mu\text{mg}$ (from(1))
$\Rightarrow\text{a}=\mu\text{g}$
$\Rightarrow4=\mu\text{g}$
$\Rightarrow\mu=\frac{4}{\text{g}}=\frac{4}{10}=0.4$
The co-efficient of kinetic friction between the block and the plane is 0.4.
$=25-20=5\text{N}$
$\therefore\tan\theta=\frac{20}{15}=\frac{4}{3}$
$\Rightarrow\mu=\tan^{-1}\Big(\frac{4}{3}\Big)=53^{\circ}$
So, the block will move at an angle 53° with an 15N force.