Question

An object of mass m = 20.0 kg travels distance d = 60.0 m at constant speed on a rough horizontal surface (μk = 0.45). The force (a pull, not a push) applied to the object makes an angle θ = 30.0∘with the horizontal. Calculate:

A:The magnitude of the applied force?

B:The work done by the applied force?

C:The work done by the force of friction?

Answer 1

A)

Balance vertical force:

N + F*sin 30 = mg

N + 0.5*F = 20.0 Kg * 9.8 m/s^2

N + 0.5*F = 196

N = 196 - 0.5 F

frictional force,

f = miuk*N

= 0.45*(196 - 0.5 F)

= 88.2 - 0.225*F

now balance horizontal force:

F*cos 30 = f

F*0.866 = 88.2 - 0.225*F

1.091*F = 88.2

F = 80.8 N

Answer: 80.8 N

B)

W = F*d*cos angle

= 80.8*60.0*cos (30)

= 80.8*60.0*0.866

= 4.20*10^3 J

Answer: 4.20*10^3 J

C)

frictional force,

f = 88.2 - 0.225*F

= 88.2 - 0.225* 80.8

= 88.2 - 18.2

= 70.0 N

W = f*d*cos angle

= 70.0*60.0*cos (180)

= 70.0*60.0*(-1)

= -4.20*10^3 J

Answer: -4.20*10^3 J