A tennis player hits a 0.1-kilogram tennis ball with her racket from the north end of court. Her racket was traveling at 65 kilometers per hour. The ball accelerated at a rate of 10 meters per second squared. The ball hits the floor on the south and of the tennis court. The floor sends the ball toward the tennis player's opponent with the same acceleration. How much force did the floor on the south and of the court exert on the ball?
To find the force exerted by the floor on the ball, we can use Newton's second law of motion, which states that the force (F) is equal to the mass (m) multiplied by the acceleration (a).
Given:
Mass of the tennis ball (m) = 0.1 kg
Acceleration of the ball (a) = 10 m/s^2
First, we need to find the initial velocity of the ball after it hits the floor on the south end of the court. To do this, we can use the equation:
Final velocity (Vf) = Initial velocity (Vi) + (acceleration * time)
Since the initial velocity (Vi) is 0 (as the ball hits the floor from a stationary position), the equation becomes:
Vf = 0 + (10 m/s^2 * time)
We can rearrange this equation to solve for time:
time = Vf / (10 m/s^2)
Given that the racket was traveling at 65 kilometers per hour, we need to convert it to meters per second:
65 km/h = 65,000 m / 3600 s = 18.06 m/s (approximately)
Plugging in this value for Vf:
time = 18.06 m/s / (10 m/s^2) = 1.806 seconds
Now we can find the force exerted by the floor using Newton's second law:
Force (F) = mass (m) * acceleration (a)
F = 0.1 kg * 10 m/s^2
F = 1 N
Therefore, the floor on the south end of the court exerts a force of 1 Newton on the tennis ball.