Question

We found that Dwight Howard could move horizontally with an acceleration of 21.6 ft/s2. What is the magnitude of the net force that Howard experiences at this horizontal acceleration?

Answer 1

The magnitude of the net force that Dwight Howard experiences at a horizontal acceleration of 21.6 ft/s² is 658 Newtons.

Step-by-step explanation:

The magnitude of the net force that Dwight Howard experiences at a horizontal acceleration of 21.6 ft/s² can be calculated using Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, since we are given the horizontal acceleration (a), we can assume the mass (m) of Dwight Howard is constant. Therefore, the magnitude of the net force can be calculated by multiplying the mass of Dwight Howard by the horizontal acceleration.

Mathematically, the formula can be written as: F = m x a.

Let's assume the mass of Dwight Howard is 100 kg. Converting the horizontal acceleration from ft/s² to m/s², we get 6.58 m/s². Substituting the values into the equation, we have: F = 100 kg x 6.58 m/s² = 658 N.

Therefore, the magnitude of the net force that Dwight Howard experiences at a horizontal acceleration of 21.6 ft/s² is 658 Newtons.