A crate of mass 13.4 kg is pulled up a rough incline with an initial speed of 1.

Question

A crate of mass 13.4 kg is pulled up a rough incline with an initial speed of 1.7G m/s. The pulling force is 117 N parallel to the incline, which makes an angle of 19.2degree with the horizontal. The coefficient of kinetic friction is 0.315 and the crate is pulled a distance of 4.G5 m. The acceleration of gravity is 9.8 m/s2 . How much work is done by gravity? Answer in units of J. How much mechanical energy is lost due to friction? Answer in units of J. How much work is done by the 117 N force? Answer in units of J. What is the change in kinetic energy of the crate? Answer in units of J. What is the speed of the crate after being pulled 4.65 m? Answer in units of m/s.

Explanation / Answer

given data:

mass of the crate m = 13.4 kg

distance d = 4.65 m

v0 = 1.76 m/s

F = 117 N and k = 0.315

from the defination of work done, w = fcas*s

(a) the workdone by gravity,

w = mgh

w = mg(d sin19.2)

    = 13.4*9.8*1.529

     = 200.818 J

therefore the work done by the gravity, w = 200.818 J.

(2) the machanical energy lost due to friction is

W = k(mgcas)d

here = 180 degree

w = 0.315 (13.4*9.8*-1)*4.65

    = -192.35 J

therefore the mechanical energy lost is -192.35 J.

(3)

W = Fnet. d

   here Fnet = F - mg(sin+kcas)

        Fnet  = 117 - 13.4*9.8(0.328+0.2974)

        Fnet = 117 - 82.137   = 34.862 N

therefore work done w = 34.862*4.65

                                 = 162.109 J

(4) the change in K.E = work done

K.E = 162.109 J

(5)

according to work energy theorm ,

K.Ef - K.Ei = W

K.Ef = w + 1/2 m Vi^2

1/2 m Vf^2 = 162.109 + 20.75

vf = 5.224 m/s

 the final speed of the crate is 5.22 m/s.     

 

   

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