Question
A tennis player hits a 0.1 kg 10 is ball with her bracket from the north end of a courts her record was traveling at 65 km/h The ball accelerated at a rate of 10 m per second squared the ball hits the floor on the south end of the tennis court The four since the ball towards the tennis player opponent with the same acceleration How much horse did the floor on the south end of the court exerts on the ball A 7.5 N B 6.5 N C 0.01 N D 1 N
Answers
GPT-4o mini
To find the force that the floor exerts on the ball, we need to apply Newton's second law of motion, which states:
\[ F = m \cdot a \]
where:
- \( F \) is the net force,
- \( m \) is the mass, and
- \( a \) is the acceleration.
In this case:
- Mass of the ball, \( m = 0.1 \) kg
- Acceleration (given) towards the opponent, \( a = 10 \) m/s²
Now, substituting the values into the formula:
\[ F = 0.1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 1 \, \text{N} \]
The ball experiences an upward normal force from the floor, which acts against its weight. The weight of the ball is calculated as:
\[ \text{Weight} = m \cdot g \]
where \( g \approx 9.8 \, \text{m/s}^2 \).
Calculating the weight:
\[ \text{Weight} = 0.1 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 0.98 \, \text{N} \]
Now, when the ball is hit and accelerates downwards, the net force exerted by the floor can be calculated. The force of the floor must overcome the weight of the ball and provide the required upward acceleration (which we can treat as going down with equal magnitude):
Net force \( F_{net} \):
\[ F_{net} = \text{Force by floor} - \text{Weight} \]
\[ F_{net} = F_{floor} - 0.98 \]
Since we already found \( F_{net} = 1 \, \text{N} \):
\[ 1 = F_{floor} - 0.98 \]
\[ F_{floor} = 1 + 0.98 = 1.98 \, \text{N} \]
However, since the floor is exerting an upward force while the ball accelerates towards the ground, its effective upward force must account for the total downward force created. Thus, the floor must apply a force of at least \( 0.98 \, \text{N} \) plus \( 1 \, \text{N} \).
In summary, considering only what option is closest to our answer of \( 1 \, \text{N} \), the closest available option from the choices provided is:
**D) 1 N.**
\[ F = m \cdot a \]
where:
- \( F \) is the net force,
- \( m \) is the mass, and
- \( a \) is the acceleration.
In this case:
- Mass of the ball, \( m = 0.1 \) kg
- Acceleration (given) towards the opponent, \( a = 10 \) m/s²
Now, substituting the values into the formula:
\[ F = 0.1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 1 \, \text{N} \]
The ball experiences an upward normal force from the floor, which acts against its weight. The weight of the ball is calculated as:
\[ \text{Weight} = m \cdot g \]
where \( g \approx 9.8 \, \text{m/s}^2 \).
Calculating the weight:
\[ \text{Weight} = 0.1 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 0.98 \, \text{N} \]
Now, when the ball is hit and accelerates downwards, the net force exerted by the floor can be calculated. The force of the floor must overcome the weight of the ball and provide the required upward acceleration (which we can treat as going down with equal magnitude):
Net force \( F_{net} \):
\[ F_{net} = \text{Force by floor} - \text{Weight} \]
\[ F_{net} = F_{floor} - 0.98 \]
Since we already found \( F_{net} = 1 \, \text{N} \):
\[ 1 = F_{floor} - 0.98 \]
\[ F_{floor} = 1 + 0.98 = 1.98 \, \text{N} \]
However, since the floor is exerting an upward force while the ball accelerates towards the ground, its effective upward force must account for the total downward force created. Thus, the floor must apply a force of at least \( 0.98 \, \text{N} \) plus \( 1 \, \text{N} \).
In summary, considering only what option is closest to our answer of \( 1 \, \text{N} \), the closest available option from the choices provided is:
**D) 1 N.**