PLEASE ANSWER CORRECTLY THE CORRECT ANSWER IS 123.93 A mass of 18.1 kg is placed

Question

PLEASE ANSWER CORRECTLY THE CORRECT ANSWER IS 123.93

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

work done by all forces = change in KE

Work by F1 + Work by F2 + Work by F3 + Work by Friction = change in KE

N = mg + F3 + F2sin(37.4) -F1sin(37.1) = 18.1*9.8 + 24 + 22*0.60 - F1*0.603

F1Cos(37.1) + 22*cos(37.4) + 0 + 0.72 ( 18.1*9.8 + 24 + 22*0.60 - F1*0.603 )*5 = 77

F1*0.797 + 22*0.794 -  0.72 ( 18.1*9.8 + 24 + 22*0.60 - F1*0.603 )*5 = 77

F1*1.23 -137 = 15.4

=> F1 = 123.9 N

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