Question

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

Answer 1

work done by tension = T* d = 49*5.9 = 289.1 J

work done by friction = f*d = $\mu_k$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