Laws Of Motion Question 11
Question 11 - 2024 (30 Jan Shift 2)
A block of mass $m$ is placed on a surface having vertical cross section given by $y=x^{2} / 4$. If coefficient of friction is 0.5 , the maximum height above the ground at which block can be placed without slipping is:
(1) $1 / 4 \mathrm{~m}$
(2) $1 / 2 \mathrm{~m}$
(3) $1 / 6 \mathrm{~m}$
(4) $1 / 3 \mathrm{~m}$
Show Answer
Answer: (1)
Solution:
$\frac{d y}{d x}=\tan \theta=\frac{x}{2}=\mu=\frac{1}{2}$
$x=1, y=1 / 4$
