Laws Of Motion Question 16
Question 16 - 2024 (31 Jan Shift 2)
A block of mass $5 kg$ 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 $g=10 m / s^{2}$ ]
[We changed options. In official NTA paper no option was correct.]
(1) $25 \sqrt{3} N$
(2) $5 \sqrt{3} N$
(3) $\frac{5 \sqrt{3}}{2} N$
(4) $10 N$
Show Answer
Answer: (2)
Solution:
$(5 \sqrt{3} N)$
$f _K=\mu mg \cos \theta$
$=0.1 \times \frac{50 \times \sqrt{3}}{2}$
$=2.5 \sqrt{3} N$
$F _1=m g \sin \theta+f _K$
