Boxes are moved on a conveyor belt from where they are filled to the packing station 11.3m away. The belt is initially stationary and must finish with zero speed. The most rapid transit is accomplished if the belt accelerates for half the distance, then decelerates for the final half of the trip. If the coefficient of static friction between a box and the belt is 0.50, what is the minimum transit time for each box?

Question

The answer is 3.04 s, but how do I get there??

Answer 1

Force=ma
mu*mg=ma
a=mu*g
well, to know the time, lets figure average velocity..for the 1/2 trip,

vf^2=vi^2+2ad=2*mu*g*11.3/2
vf= sqrt (mu*g*11.3)
and average velociyt= 1/2 Vf=1/2 sqrt(above).

That will be the avg velociyt fodr the second half also.

time=distance/avgvel=2*11.3/sqrt(above)