A mass of 17 kg is placed on a horizontal surface with a coefficient of kinetic

Question

A mass of 17 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 33 Newtons and is applied 34.8 degrees below the horizontal, the magnitude of force 3 is 36 Newtons applied in the negative y direction, and force 1 is unknown, but is applied at 34.2 degrees above the horizontal. These forces overcome static friction and the mass moves 7 meters parallel to the surface in the positive x direction. As it slides 7 meters in the positive x direction, the kinetic energy increases by 86 Joules. What is the magnitude of force 1 in Newtons?

Explanation / Answer

Here ,

F2 = 33 * ( cos(34.8) i - j * sin(34.8)) N

F1 = F1 * (cos(34.2) i + j * sin(34.2))

F3 = -36 j N

Now , as work done = net force * distance

86 = 7 * Fx

Fx = 12.3 N

let the friction is f

F1 * cos(34.2) + 33 * cos(34.8) - f = 12.3 -----(1)

let the normal force is N

u *N = f

0.51 * (36 + 33 * sin(34.8) + 17 * 9.8 - F1 * sin(34.2) ) = f -----(2)

solving 1 and 2

F1 = 88.1 N

f = 87.7 N

the force F1 is 88.1 N

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