A tennis player hits a 0.1-kilogram tennis ball with her racket from the north end of a 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 end 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 end of the court exert on the ball?

Question

(1 point)
A.) 7.5 N
B.) 1 N
C.) 0.01 N
D.) 6.5 N

Answer 1

To solve this problem, we can use Newton's second law, which states that force equals mass times acceleration (F = ma).

First, we need to convert the racket's velocity from kilometers per hour to meters per second. We know that 1 kilometer equals 1000 meters and 1 hour equals 3600 seconds. Therefore, the racket's velocity is:

65 kilometers per hour = (65 * 1000 meters) / (3600 seconds) = 18.06 meters per second

Next, we can calculate the force exerted by the racket on the ball. We know the ball's mass (0.1 kilograms) and the racket's acceleration (which is equivalent to the ball's acceleration, since they have the same value). Using Newton's second law, we have:

Force = mass * acceleration = 0.1 kg * 10 m/s^2 = 1 N

So, the force exerted by the racket on the ball is 1 Newton.

Finally, we can determine the force exerted by the floor on the ball. Since the floor sends the ball towards the tennis player's opponent with the same acceleration, the force exerted by the floor is equal in magnitude to the force exerted by the racket, which is 1 Newton.

Therefore, the correct answer is B.) 1 N.