Laws Of Motion Question 14
Question 14 - 01 February - Shift 1
A block of mass $5 kg$ is placed at rest on a table of rough surface. Now, if a force of $30 N$ is applied in the direction parallel to surface of the table, the block slides through a distance of $50 m$ in an interval of time 10s. Coefficient of kinetic friction is (given, $g=10 ms^{-2}$ ):
(1) 0.60
(2) 0.75
(3) 0.50
(4) 0.25
Show Answer
Answer: (3)
Solution:
$S=u t+\frac{1}{2} a t^{2}$
$50=0+\frac{1}{2} \times a \times 100$
$a=1 m / s^{2}$
$F-\mu m g=m a$
$30-\mu \times 50=5 \times 1$
$50 \mu=25$
$\mu=\frac{1}{2}$
