# Workign with Boxes going down a belt

Boxes are transported by a conveyor belt with a velocity v_0 to a fixed incline at A where they slide and eventually fall off at B. Knowing that u_k=0.40, determine the velocity of the conveyor belt if the boses leave the incline at B with a velocity of 2m/s.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

we will use the conservation of energy here,

the initial energy of the box E = Kinetic+ Potential

= (1/2)mv^2 + mgh

from the figure, h = 6*sin15 = 1.55

final enrgy = (1/2) m U^2

where u = 2 m/s our final velocity

thus some enrgy is lost in transition and it is used for doing work

againist the force of friction.

Frictional force = mu * normal force = 0.4 * m g cos15, from figure

workdone againist this force = force * distance = 6 * 0.4 * m g cos 15

now we can write, (1/2)mv^2 + mgh - (1/2) m U^2 = 6 * 0.4 * m g cos 15

cancel m throughout and multiply by 2 throughout

V^2 = - 2* 9.8 * 1.55 + 2^2 + 12 * 0.4 * 9.8* cos 15

= -30.38 + 4 + 45.44

or V = 4.36 m/sec

see the attached figure also.

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