Question

Factories often use machines, like conveyor belts, to move objects to different places inside the building. In one factory, 7.9 kg boxes are dropped from a height of 1.5 m onto one end of a 7.7 m long conveyor belt without bouncing. The speed of the belt is 1.3 m/s. The coefficient of friction between the box and the belt is 0.4. From the time the box is released, how long does it take for the box to reach the other end of the conveyor belt?

Answer 1

initial speed of the box is 0 m/s.

normal force=weight of the box=7.9*9.8=77.42 N

friction force=friction coefficient*normal force=30.968 N

acceleration due to friction=friction force/mass=3.92 m/s^2

friction force will accelerate the mass till it reaches speed of the conveyor i.e. 1.3 m/s

let time taken for achieving the same is t seconds.

then 0+3.92*t=1.3

==>t=0.33163 seconds

distance travelled by the box by that time=initial speed*t+0.5*acceleration*t^2

=0*0.33163+0.5*3.92*0.33163^2=0.21556 m

rest of the distance is covered at a steady speed of 1.3 m/s

time taken for reaching the end of the conveyor belt=(distance to be covered)/speed=(7.7-0.21556)/(1.3)=5.7572 seconds

hence total time taken to reach the other end of the conveyor belt=0.33163+5.7572=6.08883 seconds