Question

A ware house worker exerts a constant horizontal force of magnitude $85N$ on a $40kg$ box that is initally at rest on the floor of the ware house. After moving a distance of $2m$ the speed of the box is $1m{s}^{-1}$. The coefficent of friction between the box and the ware house floor is $(g=10m{s}^{-2})$

• A. $5/16$
• B. $3/16$
• C. $19/80$
• D. $0.25/16$

Answer 1

Answer: (B)

$3/16$

Given:-

v$=1m/s$

u$=0m/s$

f$=85N$

m=$40kg$

s$=2m$

g$=10m/s^2$

From v,u & s, we can calculate a as:-

$v^2-u^2=2as$

$\Rightarrow (1{)}^2-(0{)}^2=2\left(a\right)\left(2\right)$

$\Rightarrow a=\frac{1}{4}m/s^2$

Now, $ma=F-f$ [F:- Force applied, f:- Frictional force]

$\Rightarrow 40(\frac{1}{4}=85-{\mu }_k(40\left)\right(10)$

$\Rightarrow 10=85-{\mu }_k\left(400\right)$

$\Rightarrow {\mu }_k=\frac{75}{400}=\frac{15}{80}=\frac{3}{16}$