Laws Of Motion Question 16
Question 16 - 2024 (31 Jan Shift 2)
A block of mass $5 \mathrm{~kg}$ is placed on a rough inclined surface as shown in the figure.
If $\vec{F}{1}$ is the force required to just move the block up the inclined plane and $\vec{F}{2}$ is the force required to just prevent the block from sliding down, then the value of $\left|\vec{F}{1}\right|-\left|\vec{F}{2}\right|$ is : [Use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ]
[We changed options. In official NTA paper no option was correct.]
(1) $25 \sqrt{3} \mathrm{~N}$
(2) $5 \sqrt{3} \mathrm{~N}$
(3) $\frac{5 \sqrt{3}}{2} \mathrm{~N}$
(4) $10 \mathrm{~N}$
Show Answer
Answer: (2)
Solution:
$(5 \sqrt{3} N)$
$\mathrm{f}_{\mathrm{K}}=\mu \mathrm{mg} \cos \theta$
$=0.1 \times \frac{50 \times \sqrt{3}}{2}$
$=2.5 \sqrt{3} \mathrm{~N}$
$F_{1}=m g \sin \theta+f_{K}$
