JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 11)
A steel block of 10 kg rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration 0.2 m/s2. The normal reaction R' by the floor if mass of the iron cylinders are equal and of 20 kg each, is ____________ N. [Take g = 10 m/s2 and $$\mu$$s = 0.2]
_20th_July_Morning_Shift_en_11_1.png)
686
684
714
716
Explanation
The force equation in vertical direction is
Mg $$-$$ R = Ma
_20th_July_Morning_Shift_en_11_2.png)
where,
M = collective mass of block and all three iron cylinders = 10 + 3 $$\times$$ 20 = 70 kg
a = acceleration of block = 0.2 ms$$-$$2
g = 10 ms$$-$$2 and $$\mu$$s = 0.2
and R = normal reaction
Force along vertical axis Mg $$-$$ R = Ma
$$\therefore$$ 70g $$-$$ R = 70 $$\times$$ 0.2
$$\Rightarrow$$ R = 70 $$\times$$ 10 $$-$$ 14
= 700 $$-$$ 14 = 686 N
Mg $$-$$ R = Ma
where,
M = collective mass of block and all three iron cylinders = 10 + 3 $$\times$$ 20 = 70 kg
a = acceleration of block = 0.2 ms$$-$$2
g = 10 ms$$-$$2 and $$\mu$$s = 0.2
and R = normal reaction
Force along vertical axis Mg $$-$$ R = Ma
$$\therefore$$ 70g $$-$$ R = 70 $$\times$$ 0.2
$$\Rightarrow$$ R = 70 $$\times$$ 10 $$-$$ 14
= 700 $$-$$ 14 = 686 N
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