Question

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 $10 \,s$ Coefficient of kinetic friction is (given, $g =10 \,ms ^{-2}$ ):

• A. 0 .50
• B. $0.60$
• C. $0.75$
• D. $0.25$

Answer 1

The Correct Option is A

$S=ut+\frac{1}{2}a{t}^2$
$50=0+\frac{1}{2}\times a\times 100$
$a=1m/s^2$
$F-\mu mg=ma$
$30-\mu \times 50=5\times 1$
$50\mu =25$
$\mu =\frac{1}{2}$

Answer 2

Using kinematic equation: \[ S = ut + \frac{1}{2} a t^2 \] \[ 50 = 0 + \frac{1}{2} a \times 100 \] \[ a = 1 \text{ m/s}^2 \] Force equation: \[ F - \mu mg = ma \] \[ 30 - \mu \times 50 = 5 \times 1 \] \[ 50\mu = 25 \] \[ \mu = \frac{1}{2} \]