Question

The conveyor belt is designed to transport packeage of varies weights. Each $10kg$ packege has a coefficent of kinetic friction ${\mu }_k=0.15$. If the speed of the conveyor is $5m/s$, and then it suddenly stops, determine the distance the package will slide on the belt before coming to rest.

• A. $3.8m$
• B. $2.9m$
• C. $1.4m$
• D. $8.5m$

Answer 1

The correct option is D $8.5m$

Given,

Initial kinetic energy, $K.E=\frac{1}{2}m{v}^2=\frac{1}{2}\times 10\times 5^2=125J$

Friction force, $F_r=\mu mg=0.15\times 10\times 9.8=14.7N$

Initial kinetic energy = work done by friction force

$K.E=F_{r}d$

$125J=14.7\times d$

$d=\frac{125}{14.7}=8.5m$

After $8.5m$ package comes to rest.