Laws Of Motion Ans 15

Question 15 - 01 February - Shift 2

As shown in the figure a block of mass $10 kg$ lying on a horizontal surface is pulled by a force $F$ acting at an angle $30^{\circ}$, with horizontal. For $\mu_s=0.25$, the block will just start to move for the value of $F$ : [Given $g=10 ms^{-2}$ ]

(1) $33.3 N$

(2) $25.2 N$

(3) $20 N$

(4) $35.7 N$

Show Answer

Answer: (2)

Solution:

Formula: Second Law Of Motion

$N=Mg-F Sin 30^{\circ}$

$=mg-\frac{F}{2}=100-\frac{F}{2}=\frac{200-F}{2}$

$F Cos 30^{\circ}=\mu N$

$\sqrt{3} \frac{F}{2}=0.25 \times(\frac{200-F}{2})$

$4 \sqrt{3} F=200-F$

$F=\frac{200}{4 \sqrt{3}+1}=25.22$