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
$N=Mg-F Sin 30^{\circ}$
$=mg-\frac{F}{2}=100-\frac{F}{2}=\frac{200-F}{2}$
$F Cos 30^{\circ}=\mu N$
$\sqrt{3} \frac{F}{2}=0.25 \times(\frac{200-F}{2})$
$4 \sqrt{3} F=200-F$
$F=\frac{200}{4 \sqrt{3}+1}=25.22$
