What is the normal force when a 40 kg person jumps and is accelerating at 3 m/s2 upward while in contact with the ground?
force=m(g-3)
To find the normal force acting on a person, we need to consider Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the person, with a mass of 40 kg, is accelerating upward at 3 m/s^2.
The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the ground is exerting a normal force on the person to counteract their weight.
The weight of an object can be calculated by multiplying its mass by the acceleration due to gravity. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2.
Weight = Mass * Acceleration due to gravity
Weight = 40 kg * 9.8 m/s^2
Weight = 392 N
Since the person is accelerating upward with a force of 3 m/s^2, we can calculate the net force acting on them.
Net Force = Mass * Acceleration
Net Force = 40 kg * 3 m/s^2
Net Force = 120 N
Since the normal force counteracts the weight and the net force, and they are in opposite directions, the normal force can be calculated as:
Normal Force = Weight + Net Force
Normal Force = 392 N + 120 N
Normal Force = 512 N
Therefore, the normal force acting on the person when they are accelerating at 3 m/s^2 upward while in contact with the ground is 512 N.