A mass m = 14 kg is pulled along a horizontal floor, with a coefficient of kinetic friction μ_{k} = 0.12, for a distance d = 5.9 m. Then the mass is continued to be pulled up a frictionless incline that makes an angle θ = 38° with the horizontal. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of θ = 38° (thus on the incline it is parallel to the surface) and has a tension T = 49 N.
1)What is the work done by tension before the block gets to the incline?
______________J
2)What is the work done by friction as the block slides on the flat horizontal surface?
______________J
3)What is the speed of the block right before it begins to travel up the incline?
_____________m/s
4)How far up the incline does the block travel before coming to rest?
_______________m
5)What is the work done by gravity as it comes to rest?
______________J
6)During the entire process, the net work done on the block is:
positive
negative
zero
work done by tension = T* d = 49*5.9 = 289.1 J
work done by friction = f*d = mg*d = 0.12*14*9.8*5.9 = 97.1376 J
Wtension - Wfriction = 1/2m*v^2
289.1 - 97.1376 = 0.5*14*v^2
V = 5.23 m/s
V = 5.23 m/s just before inclined we want to it stop so
Wt - Wmg = KEf-KEi = 0-1/2.*14*5.23^2
49*d - mg*d*sin38 = - 0.5*15*5.23^2
49*d - 14*9.8*d*sin38 = - 0.5*15*5.23^2
d = 5.78 m ............answer
work done by gravity = - 14*9.8*d*sin38 = - 14*9.8*5.78*sin38 = -488.229 J
6 ) total work done = negative
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