Kinetic Energy, Work Done, and Inclines

[orbits]

UPS delivery man is loading his truck by shoving the parcels up a ramp. The ramp is inclined at an angle of 30 degrees above the horizontal. He sends a parcel of mass 15 kg, up this ramp. the coefficient of sliding friction is .10 and gravitational force is class standard (10).

A The magnitude of the kinetic friction acting on the parcel is?
B He sends parcel up ramp with initial Kinetic Energy, E, so that by the time, the parcels up the ramp, it has 0 KE. Ramp is length 1.0m.
C To give this parcel this initial KE, he most do WORK on the parcel. Assuming he pushed it a short distance of .75m what is the average force he needs to apply. Ignore friction for this part.


The work I have done is:

A
Y Components: N-MG Cos 30 --> N=150 Cos 30 --> N=129.9
Friction = \muk * N --> .10 * 129.9 ---> Magnitude =12.99

B

1/2 M V_o²+ [STRIKE]MGH[/STRIKE]= [STRIKE]1/2 M V_f²[/STRIKE] + MGH

1/2 M V_o² = MGH


H= Sin30 -->.5


1/2 (15) (V)²= 75
V=3.16

So initial KE = 75 ?


C

Is related to B so if that's wrong this is wrong.

Work = Force * Displacement
Work = KE

KE: 75 / Disp: .75 = Force: 100N


Thanks for your help.

 

[denverdoc]

Part a looks fine. Part B is missing the effects of frictional dissipation which will rob some of the PE.

C is done by ignoring friction but asks that you compute force from the work done and the displacement over which force is appplied. Assuming he pushes parallel to the ramp--this will be F*d=energy required to get to top. That help?

 

[orbits]

So for B:

Energy - Friction= 75-12.99 = 62.01For C:

Would I use the new energy 62.01/ (.75) = 86.68N