Question

A block is gently placed on a conveyor belt moving horizontally at constant speed. After $t=4s$, the velocity of the block becomes equal to the velocity of the belt. If the coefficient of friction between the block and the belt is $\mu =0.2$, then the velocity of the conveyor belt is?

• A. $2m{s}^{-1}$
• B. $4m{s}^{-1}$
• C. $64m{s}^{-1}$
• D. $8m{s}^{-1}$

Answer 1

The correct option is D $8m{s}^{-1}$

$umg=\frac{m(v_f-v_s)}{t}$

$0.2\times 10=\frac{V-0}{4}=2\times 4=v$

$v=8m/s$