# Question #6e844

Dec 3, 2016

Given

$m \to \text{mass of the block} = 9.25 k g$

$F \to \text{horizontal force applied on the block} = 55 N$

${\mu}_{k} \to \text{coefficient of kinetic friction} = 0.175$

${F}_{\text{fric"-> "frictional force resisting the block}}$
$\text{ "" } = {\mu}_{k} m g = 0.175 \times 9.25 \times 9.8 N = 15.86375 N$

So net force acting on the block

${F}_{n} = F - {F}_{\text{fric}} = \left(55 - 15.86375\right) = 39.13625 N$

So acceleration produced

$a = {F}_{n} / m = \frac{39.13625 N}{9.25 k g} \approx 4.23 m {s}^{-} 2$

If time taken to cover the distance $s = 3 m$ be t sec then by using equation of kinematics we can write

$S = u \times t + \frac{1}{2} a {t}^{2}$

$\implies 3 = 0 \times t + \frac{1}{2} \times 4.23 \times {t}^{2}$

$t = \sqrt{\frac{6}{4.23}} s = \approx 1.19 s$