JEE Main 2019 — Laws Of Motion Question with Solution

N = mg + 20 sin30^o

    = 50 + $$20\times \frac{1}{2}$$

    = 60 N

Limition friction, f_L = $$\mu$$N = 0.2 $$\times$$ 60 = 12 N

And horizontal force(F_x) = 20 cos30^o = 10$$\sqrt{3}$$ = 17.32 N

As F_x > f_L, then the block will move with acceleration a_A.

a_A = $$\frac{F_x-f_L}{m_A}$$ = $$\frac{17.32-12}{5}$$ = $$\frac{5.32}{5}$$

N + 20 sin30^o = mg

N = mg - 20 sin30^o

    = 50 - 10 = 40 N

Limition friction, f_L = $$\mu$$N = 0.2 $$\times$$ 40 = 8 N

And horizontal force(F_x) = 20 cos30^o = 10$$\sqrt{3}$$ = 17.32 N

As F_x > f_L, then the block will move with acceleration a_B.

a_B = $$\frac{F_x-f_L}{m_B}$$ = $$\frac{17.32-8}{5}$$ = $$\frac{9.32}{5}$$

Difference between acceleration a_A - a_B = $$\frac{9.32}{5}-\frac{5.32}{5}$$ = $$\frac{4}{5}$$ = 0.8 m/s²