a 2.4*10^2 N force is pulling an 85 kg refrigerator across a horizontal surface. the force acts at an angle of 20.0 above the surface. the coefficient of kinetic friction is .200, and the refrigerator moves a distance of 8.00 m. find (a) the work dont by the pulling force, and (b) the work done by the kinetic frictional force

Respuesta :

Answer:

a) The work done by pulling force, W = 1804.21 joules

b) The work done by the kinetic frictional force is, w = 1332.8 J

Explanation:

Given,

The pulling force, Fₐ = 2.4 x 10² newton

The mass of the refrigerator, m = 85 Kg

The angle of pulling force acting to the surface, ∅ = 20°

The coefficient of kinetic friction is, μₓ = 0.200

The distance covered by the refrigerator, s = 8 m

The force acting in the direction parallel to the surface is given by

                                      F = Fₐ cos∅

∴                                     F = 2.4 x 10² x cos 20°

                                         =225.53 N

The work done by the pulling force is given by

                                  W = F · S

                                       = 225.53 N x 8 m

                                       = 1804.21 joules

The work done by the pulling force, W = 1804.21 joules

The kinetic friction force is given by the formula

                                  Fₓ = μₓ · η

Where

                              μₓ - coeficient of kinetic frictoin

                              η - Normal force, (mg)

∴                             Fₓ = 0.200 x 85 x 9.8

                                   =  166.6 N

The work done by the kinetic frictional force is given by

                                 w = Fₓ · S

                                     = 166.6 N x 8 m

                                     = 1332.8 J

Hence, the work done by the kinetic frictional force is, w = 1332.8 J

Answer:

a) 1800 Joules

b) -1201.47Joules

Explanation:

Workdone by pulling force= F(cos angle)× s

Where s = displacement

W= 240N(cos20°)×8m

W= 1800 Joules

b) The frictional force will be opposite the displacement, therefore the angle will be 180°

Workdone by kinetic frictional force Fs=Fn × coefficient of kinetic friction

W= [(85×9.8) - 240sin20°) ]× [0.200cos180×8]

W= (833-82.08) × (-1.6)

W= 750.92×(-1.6)

W=-1201.47Joules

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