Hello, I have tried to attack this problem using thework-energy equation: U1-2=T

Question

Hello, I have tried to attack this problem using thework-energy equation:
U1-2=T2-T1=?T-or- T1+U1-2=T2 but canonly get as far as a couple of steps into it and everything lookswrong...is this the wrong equation to use? If not, doesanyone know which one to use? If I am on the righttrack, can I get a little help on the steps to use withthe work-energy eqn? I'm lost and pretty confusedwith this one. Thanks for looking.

In the design of a conveyor belt system, small metalblocks are discharged with a velocity of 0.4m/s onto a ramp by theupper conveyor belt shown. If the coefficient of kineticfriction between the blocks and the ramp is 0.30, calculate theangle ? which the ramp must make with the horizontal so thatthe blocks will transfer without slipping to the lower conveyorbelt moving at the speed of 0.14m/s. (given answer in book:?=16.62o)


how did u get the answer

Explanation / Answer

ok. So Friction is mg*cos(theta)*Uk. F*D=Work. Work=change in KE mg*cos(theta)*Uk*distance=1/2*m*(vf^2-vi^2) -9.8*cos(theta)*.3*d=1/2*m*(.14^2-.4^2) In order for theta to be 16.62, I'm guessing your distance is 0.07 meters. I need distance though.

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