Question

A long conveyor belt is designed to transport packages of various weights. Each $20kg$ package has a coefficient of kinetic friction $0.25$. If the speed of the conveyor belt is $10m/s$ and then it suddenly stops, the distance the package will slide before coming to rest is

• A. $6\text{m}$
• B. $10\text{m}$
• C. $15\text{m}$
• D. $20\text{m}$

Answer 1

The correct option is D $20\text{m}$

When the conveyor belt is suddenly stopped, the block is moving with velocity $10m/s$. Let $a$ be the deceleration of the block (due to kinetic friiction between the block and the belt)

The FBD of the package can be given by


From the FBD, we have
$N=20g=200\text{N}$
$f_k={\mu }_{k}N=0.25\times 200=50\text{N}$
$a=\frac{f_k}{m}=\frac{50}{20}=2.5m/s^2$

Hence, using 3rd equation of motion,
Distance travelled before sliding stops is
$s=\frac{v^2}{2a}=\frac{(10{)}^2}{2\times 2.5}=\frac{100}{5}=20\text{m}$