A mass of 20.5 kg is placed on a horizontal surface with a coefficient of kineti

Question

A mass of 20.5 kg is placed on a horizontal surface with a coefficient of kinetic friction of 0.51. Three forces are applied to it as shown with force 2 having a magnitude of 35 Newtons and is applied 30.1 degrees below the horizontal, the magnitude of force 3 is 25 Newtons applied in the negative y direction, and force 1 is unknown, but is applied at 36 degrees above the horizontal. These forces overcome static friction and the mass moves 8 meters parallel to the surface in the positive x direction.   As it slides 8 meters in the positive x direction, the kinetic energy increases by 71 Joules. What is the magnitude of force 1 in Newtons?

Explanation / Answer

apply newtons II law in vertical direction

N + F1 sin 30.1 = F2 sin 90 + 25 sin 36 + mg

N + F sin 30.6 = 35 + 14.69 + 20.5 * 9.8

N = 250.59 - 0.51 F1 -----------------------------1

Frictional force Fk = ukN = 0.51 * (250.59 - 0.51 F1)

Fk = 127.8 -0.26 F1

Net Horizontal force Fnet = Fcos 30.6 + 35 - 127.8 -0.51 F

Fnet = 1.36 F - 92.8

as W = F dx

1.36 F1 -92.8 = 71/8

F1 = 74.76 N   

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