# 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?

## F is up, +

F - m g = m a

F = m(g+a) = 40 (9.81+3)

## To find the normal force acting on a person while jumping, we need to consider the forces acting on the person in equilibrium.

The normal force is the force exerted by a surface to prevent an object from falling through it.

In this case, the normal force would be equal in magnitude and opposite in direction to the gravitational force acting on the person.

The gravitational force acting on a person can be calculated using the formula:

Weight = mass × acceleration due to gravity

Weight = 40 kg × 9.8 m/s²

Weight = 392 N

Since the person is accelerating at 3 m/s² upward, the net force acting on the person can be calculated using Newton's second law:

Net force = mass × acceleration

Net force = 40 kg × 3 m/s²

Net force = 120 N

In equilibrium, the net force is equal to zero. Therefore, the normal force would be:

Normal force = Weight - Net force

Normal force = 392 N - 120 N

Normal force = 272 N

Therefore, the normal force acting on the person is 272 N.

## To calculate the normal force acting on the person while in contact with the ground, 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.

Here's how you can find the normal force:

1. Identify the known values:
- Mass of the person, m = 40 kg
- Acceleration, a = 3 m/s² (upward)

2. Determine the net force acting on the person:
- Net force (F_net) = mass (m) × acceleration (a)
- F_net = 40 kg × 3 m/s² = 120 N (upward)

3. Evaluate the normal force:
- The normal force (F_normal) is equal in magnitude but opposite in direction to the net force acting on the person.
- Therefore, the normal force is 120 N but directed downward.

So, the normal force acting on the person when they jump and accelerate at 3 m/s² upward is 120 N, directed downward.