Question

Q 75. A block lying on a long horizontal conveyor belt moving at a constant velocity receives a velocity 5 m/s relative to the ground in the direction opposite to the direction of motion of the conveyor. After t= 4 sec, the velocity of the block becomes equal to the velocity of the belt.The coefficient of friction between the block and the belt is 0.2. Then the velocity of the conveyor belt is.

Answer 1

Dear Student,


$$\begin{gathered} letthevelocityofthebeltbev \\ relativevelocityoftheblockwiththebelt=v+5 \\ solvethequestionwrtbelt. \\ u=v+5finalvelocitywrtbelt=0 \\ forceoffriction=\mu mg \\ accn=\frac{-\mu mg}{m}=-\mu g \\ finalvelocity=u+at \\ 0=v+5-\mu gt \\ v=0.2*10*4-5=8-5=3m/svelocityofbelt. \\ Regards \end{gathered}$$