Question #108322

To unload a truck a down ramp of length 8.0 m is arranged so it is downwards at an angle of

30.0° to the horizontal. Boxes of fruit and placed on the ramp and they slide down attaining a

velocity of 4.00 m/s by the time it reaches the bottom of the ramp. Determine the energy loss

due to friction while sliding down the ramp by considering the changes in the potential energy

and the kinetic energy of the box, between the top and bottom of the ramp. Hence determine a

value for the kinetic coefficient of friction

Expert's answer

Elost=mgh0.5mv2E_{lost}=mgh-0.5mv^2

Elost=m(glsin300.5(v)2)E_{lost}=m(gl\sin{30}-0.5(v)^2)

Elost=4((9.8)(8)(0.5)0.5(4)2)=125 JE_{lost}=4((9.8)(8)(0.5)-0.5(4)^2)=125\ J

μ=1254((9.8)(8)(0.5))=0.80\mu=\frac{125}{4((9.8)(8)(0.5))}=0.80


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