Question

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 players opponent with the same acceleration. how much force did the floor on the south end of the court exert on the ball? (1 point)
○ 0.01 n
○ 7.5 n
○ 1 n
○ 6.5 n

Explanation:

Step1: Recall Newton's second law

Newton's second law is \(F = ma\), where \(F\) is force, \(m\) is mass, and \(a\) is acceleration.

Step2: Identify the values of \(m\) and \(a\)

The mass \(m = 0.1\space kg\) and the acceleration \(a=10\space m/s^{2}\) (given that the floor sends the ball with the same acceleration as when it was hit).

Step3: Calculate the force

Substitute \(m = 0.1\space kg\) and \(a = 10\space m/s^{2}\) into \(F=ma\). So \(F=(0.1\space kg)\times(10\space m/s^{2})\).

Answer:

\(1\space N\) (since \(F=(0.1)\times(10)=1\))