Explanation
For 10kg block, f is frictional force
$$f_{1max} = \mu_1 \mu_2 = \mu_1 (10g) = 0.1(10 \times 10) = 10N$$
i) if $$F = 2N$$ then acting friction
$$f_1 = F = 2N$$
(A) is correct.
for 5 kg block;
$$f_{2max} = \mu (10 + 5)g = 0.3 \times 15 \times 10= 45 N$$
ii) if $$F = 30N$$ then;
$$F - F_{1max}= 10a$$
$$10a = 30 - 10 =20$$
$$a =2 m/s^2$$
(B) is correct.
iii) if $$\mu_1$$ changed to 0.5;
then $$f_{1max} = \mu_1 (10g) = 0.5 \times 10 \times 10 = 50N$$
In the case acting friction will be $$F_{2max}$$
Since,
$$F_{1max} = F_{2max}$$
Block 10kg and 5kg both move together then $$F_{min}$$ to begin is
$$F_{min} - F_{max} = 0$$
$$F_{min} = F_{2max} = 45 N$$
(c) is correct.
iv) if $$\mu = 0.1$$ , then $$f_{max} = 10 N$$ and,
$$f_{2max} = 45 N$$
$$f_{2max} > f_{1max}$$ always for any any value of $$N$$. Hence, 5kg block never move on the ground.
So, option (A), (B), (C), and (D) are correct.
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