A mass of 9 kg is resting on a horizontal, frictionless surface. Force 1 is appl

Question

A mass of 9 kg is resting on a horizontal, frictionless surface. Force 1 is applied to it at 28 degrees below the horizontal, force 2 has a magnitude of 22 N and is applied vertically downward, force 3 has a magnitude of 4 N and is applied vertically upwards, and force 4 has a magnitude of 21 N and is applied in the -x direction to the object. When these forces are applied to the object, the object moves 9 meters in the +x direction and the velocity of the object is 6 m/s in the +x direction. What is the normal force acting on the mass in Newtons?

Explanation / Answer

Using 3rd equation of motion,

Acceleration, a= v^2/(2d) = 6^2/(2*9) = 2 m/s^2

Considering equilibrium of forces along x axis

ma = F1 cos x - F4

9 x 2 = F1 cos 28 - 21

F1= 44.17 N

Normal force given by

N= mg+ F2 + F1 sin x - F3

= 9 x 9.8 + 44.17 sin 28 + 22 - 4

= 126.94 N

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Commemt in case any doubt.. good luck

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