Question

The conveyor belt is designed to transport packages of various weights. Each 10 kg package has a coefficient of kinetic friction $\mathrm{0.15.}$ If the speed of the conveyor belt is $5m/s$, and then it suddenly stops, determine the distance the package will slide before coming to rest:
(g = $9.8m/s^2$)

• A. $8.5m$
• B. $8m$
• C. $10m$
• D. $6m$

Answer 1

Answer: (A)

$8.5m$

The block is initially moving with velocity 5 m/s

when belt stops the friction force retard the motion of block

friction force

$F=\mu mg$

$F=0.15\times 10\times 9.8=14.7N$

retardation $a$ =$\frac{F}{m}$

$\frac{F}{m}=\mu g=1.47$

now,

$v^2=u^2+2as0=5^2-2\times 1.47\times ss=8.5m$

Option A is correct.