JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 8)
A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F = 20 N, making an
angle of 30o with the horizontal, as shown in the figures. The coefficient of friction between the block and
floor is $$\mu $$ = 0.2. The difference between the accelerations of the blocks, in case (B) and case (A) will be :
(g = 10 ms–2)
_12th_April_Evening_Slot_en_8_1.png)
3.2 ms–2
0.8 ms–2
0 ms–2
0.4 ms–2
Explanation
_12th_April_Evening_Slot_en_8_2.png)
N = mg + 20 sin30o
= 50 + $$20 \times {1 \over 2}$$
= 60 N
Limition friction, fL = $$\mu $$N = 0.2 $$ \times $$ 60 = 12 N
And horizontal force(Fx) = 20 cos30o = 10$$\sqrt 3 $$ = 17.32 N
As Fx > fL, then the block will move with acceleration aA.
aA = $${{{F_x} - {f_L}} \over {{m_A}}}$$ = $${{17.32 - 12} \over 5}$$ = $${{5.32} \over 5}$$
_12th_April_Evening_Slot_en_8_3.png)
N + 20 sin30o = mg
N = mg - 20 sin30o
= 50 - 10 = 40 N
Limition friction, fL = $$\mu $$N = 0.2 $$ \times $$ 40 = 8 N
And horizontal force(Fx) = 20 cos30o = 10$$\sqrt 3 $$ = 17.32 N
As Fx > fL, then the block will move with acceleration aB.
aB = $${{{F_x} - {f_L}} \over {{m_B}}}$$ = $${{17.32 - 8} \over 5}$$ = $${{9.32} \over 5}$$
Difference between acceleration aA - aB = $${{9.32} \over 5} - {{5.32} \over 5}$$ = $${4 \over 5}$$ = 0.8 m/s2
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