A conveyor belt is used to move sand in a cement factory. The conveyor is liked

Question

A conveyor belt is used to move sand in a cement factory. The conveyor is liked up at an angle of 14^th from the horizontal and the sand is moving without slipping as 7.0 m/s. The sand is to land in the middle of a mixing tub located 3.0 m below the end of the conveyor belt Write the x and y position and velocity equations for the sand while it is in the air. Fill in those quantities you can deduce from the problem statement. Determine the horizontal distance from the end of the conveyor belt to the middle of the tub What is the velocity vector for the sand as it enters the tub

Explanation / Answer

if the angle of the conveyer belt from horizontal is 14 degress and height of the tub from the ground is 3 m and it is moving at the speed of 7 m/sec. so it will gain potential energy while it's velocity will remain constant -

potential energy = mgh

P = 3gh

K = 0.5 mv2

K = 0.5m*49

K = 24.5 m

total energy of the sand grain of mass 'm' -

E = K + P

E = 53.9 m

when it leaves the conveyer suppose it's velocity is v' then -

0.5 mv'2 = 53.9 m

v'2 = 107.8

v' = 10.38 m/sec

horizontal component of velocity = v' cos 14 = 10.07

vertical component of velocity = v' sin 14 = 2.51

x = v' cos 14 t + at

x = 10.07 t ..........................1

y = v' sin 14 t - 9.8 t2

y = 2.52 t - 9.8 t2 ........................2

here y = - 3

t = 0.69 sec

so x = 6.94 m

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