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

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

7.5 N

1 N

0.01 N

6.5 N

Answer 1

To calculate the force exerted by the floor on the ball, we can use Newton's second law of motion, which states that force (F) equals mass (m) multiplied by acceleration (a). In this case, the mass of the ball is 0.1 kg and the acceleration is 10 m/s^2.

Therefore, the force exerted by the floor on the ball can be calculated as:

F = m x a

F = 0.1 kg x 10 m/s^2

F = 1 N

So, the floor exerts a force of 1 N on the ball.

Answer 2

To find the force exerted by the floor on the ball, we can use Newton's second law of motion:

Force = mass x acceleration

First, let's convert the mass of the tennis ball to kilograms:
1 kilogram = 1000 grams
0.1 kilograms = 100 grams

Now, let's calculate the initial velocity of the ball (v0) using the velocity of the racket (65 kilometers per hour) and converting it to meters per second:
1 kilometer = 1000 meters
1 hour = 3600 seconds

v0 = 65 kilometers per hour x (1000 meters / 1 kilometer) x (1 hour / 3600 seconds)
v0 = 65,000 meters / 3600 seconds
v0 ≈ 18.056 meters per second

Now, let's calculate the time it takes for the ball to hit the floor using the acceleration of the ball (10 meters per second squared):
We can use the kinematic equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

v = 0 (since the ball comes to rest when it hits the floor)
u = 18.056 meters per second
a = 10 meters per second squared

0 = 18.056 + 10t
t ≈ -1.806 seconds
(Note: The negative time indicates that it took approximately 1.806 seconds for the ball to hit the floor after being struck by the racket.)

Since we are concerned with the magnitude of the force exerted by the floor, we can consider only the positive time:
t ≈ 1.806 seconds

Now, let's calculate the force exerted by the floor using Newton's second law of motion:
Force = mass x acceleration

Force = 0.1 kilograms x 10 meters per second squared
Force = 1 kilogram meter per second squared (N)

Therefore, the force exerted by the floor on the south end of the court on the tennis ball is 1N.

Answer 3

To find the force exerted by the floor on the south end of the court on the tennis ball, we can use Newton's second law of motion. The formula is:

Force (F) = mass (m) x acceleration (a)

First, let's convert the speed of the racket from kilometers per hour to meters per second. Since 1 kilometer is equal to 1000 meters and 1 hour is equal to 3600 seconds, we can use the conversion factor:

65 kilometers per hour = (65 * 1000) meters / (3600 seconds)
= 18.06 meters per second (rounded to two decimal places)

Given that the mass of the ball is 0.1 kilograms, and the acceleration is 10 meters per second squared, we can now calculate the force exerted by the floor on the ball.

Force (F) = mass (m) x acceleration (a)
= 0.1 kilograms x 10 meters per second squared
= 1 Newton

Therefore, the floor exerts a force of 1 Newton on the ball. So the correct answer is 1 N.