A 2.9 kg ball strikes a wall with a velocity of 7.5 m/s to the left. The ball bounces off with a velocity of 6.3 m/s to the right. If the ball is in contact with the wall for 0.20 s, what is the constant force exerted on the ball by the wall? _________ N to the right

To find the constant force exerted on the ball by the wall, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

First, we need to calculate the acceleration of the ball. We can use the equation:

acceleration = (final velocity - initial velocity) / time

The initial velocity of the ball is -7.5 m/s (to the left) and the final velocity is 6.3 m/s (to the right). The time the ball is in contact with the wall is 0.20 s. Plugging these values into the equation, we get:

acceleration = (6.3 m/s - (-7.5 m/s)) / 0.20 s

acceleration = (6.3 m/s + 7.5 m/s) / 0.20 s

acceleration = 13.8 m/s / 0.20 s

acceleration = 69 m/s^2

Now, we can calculate the force exerted on the ball by the wall using the equation:

force = mass * acceleration

The mass of the ball is 2.9 kg, and the acceleration calculated earlier is 69 m/s^2. Plugging these values into the equation, we get:

force = 2.9 kg * 69 m/s^2

force = 200.1 N

Therefore, the constant force exerted on the ball by the wall is 200.1 N to the right.

force*time=2.9(7.5+6.3)