Question:
The coefficient of static friction between two blocks is $0.5$ and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is .......N.
$\left(\right.$ take $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
Solution:
$\mathrm{F}=3 \mathrm{a}($ For system $)$ ........................(I)
$\mathrm{fs}_{\max }=1 \mathrm{a}($ for $1 \mathrm{~kg}$ block) ..................(II)
$\mu \times 1 \times g=a$
$\Rightarrow 5=\mathrm{a}$
$\mathrm{F}=15 \mathrm{~N}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.