A mass of 18.1 kg is placed on a horizontal surface with a coefficient of kineti

Question

A mass of 18.1 kg is placed on a horizontal surface with a coefficient of kinetic friction of 0.72. Three forces are applied to it as shown with force 2 having a magnitude of 22 Newtons and is applied 37.4 degrees below the horizontal, the magnitude of force 3 is 24 Newtons applied in the negative y direction, and force 1 is unknown, but is applied at 37.1 degrees above the horizontal. These forces overcome static friction and the mass moves 5 meters parallel to the surface in the positive x direction.   As it slides 5 meters in the positive x direction, the kinetic energy increases by 77 Joules. What is the magnitude of force 1 in Newtons?

A mass of 18.1 kg is placed on a horizontal surface with a coefficient of kinetic friction of 0.72. Three forces are applied to it as shown with force 2 having a magnitude of 22 Newtons and is applied 37.4 degrees below the horizontal, the magnitude of force 3 is 24 Newtons applied in the negative y direction, and force 1 is unknown, but is applied at 37.1 degrees above the horizontal. These forces overcome static friction and the mass moves 5 meters parallel to the surface in the positive x direction.   As it slides 5 meters in the positive x direction, the kinetic energy increases by 77 Joules. What is the magnitude of force 1 in Newtons?

99.45

123.93 ± 10%

Explanation / Answer

F=f1x+f2x+F3 =F1cos37.1+22cos37.4+24cos90 = F1cos37.1+17.48

Fn-F1sin37.1-22sin37.4 -24-mg=0

Fn= 24-22sin37.4-F1sin37.1+18.1*9.8 =207.86 - F1sin37.1

Fnet= F-Ff=F-uk*Fn =(F1cos37.1+17.48) – 0.72*(24-22sin37.4-F1sin37.1)

W= Fnet*d

77=[(F1cos37.1+17.48) – 0.72*(207.86 - F1sin37.1)]*5           => F1= 119.8 N

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