Answer :
To determine the force applied to bring the ball to a stop in 2 seconds, we need to follow a series of steps involving basic principles of kinematics and Newton’s second law of motion. Let's go through this step-by-step.
1. Identify the given information:
- Initial velocity ([tex]\(v_i\)[/tex]): [tex]\(20 \, \text{m/s}\)[/tex]
- Final velocity ([tex]\(v_f\)[/tex]): [tex]\(0 \, \text{m/s}\)[/tex]
- Time ([tex]\(t\)[/tex]): [tex]\(2 \, \text{s}\)[/tex]
- Mass ([tex]\(m\)[/tex]): [tex]\(200 \, \text{g}\)[/tex] which is equivalent to [tex]\(200 / 1000 = 0.2 \, \text{kg}\)[/tex]
2. Calculate the acceleration ([tex]\(a\)[/tex]):
Using the formula for acceleration:
[tex]\[ a = \frac{v_f - v_i}{t} \][/tex]
We substitute the values into the formula:
[tex]\[ a = \frac{0 - 20}{2} = \frac{-20}{2} = -10 \, \text{m/s}^2 \][/tex]
This negative sign indicates that it is a deceleration.
3. Calculate the force ([tex]\(F\)[/tex]):
Using Newton’s second law of motion:
[tex]\[ F = m \cdot a \][/tex]
Substituting the values:
[tex]\[ F = 0.2 \, \text{kg} \cdot (-10 \, \text{m/s}^2) = -2 \, \text{N} \][/tex]
The negative sign indicates that the force is applied in the direction opposite to the direction of motion.
In conclusion, the force applied by the player on the ball to bring it to a stop in 2 seconds is [tex]\(-2 \, \text{N}\)[/tex].
1. Identify the given information:
- Initial velocity ([tex]\(v_i\)[/tex]): [tex]\(20 \, \text{m/s}\)[/tex]
- Final velocity ([tex]\(v_f\)[/tex]): [tex]\(0 \, \text{m/s}\)[/tex]
- Time ([tex]\(t\)[/tex]): [tex]\(2 \, \text{s}\)[/tex]
- Mass ([tex]\(m\)[/tex]): [tex]\(200 \, \text{g}\)[/tex] which is equivalent to [tex]\(200 / 1000 = 0.2 \, \text{kg}\)[/tex]
2. Calculate the acceleration ([tex]\(a\)[/tex]):
Using the formula for acceleration:
[tex]\[ a = \frac{v_f - v_i}{t} \][/tex]
We substitute the values into the formula:
[tex]\[ a = \frac{0 - 20}{2} = \frac{-20}{2} = -10 \, \text{m/s}^2 \][/tex]
This negative sign indicates that it is a deceleration.
3. Calculate the force ([tex]\(F\)[/tex]):
Using Newton’s second law of motion:
[tex]\[ F = m \cdot a \][/tex]
Substituting the values:
[tex]\[ F = 0.2 \, \text{kg} \cdot (-10 \, \text{m/s}^2) = -2 \, \text{N} \][/tex]
The negative sign indicates that the force is applied in the direction opposite to the direction of motion.
In conclusion, the force applied by the player on the ball to bring it to a stop in 2 seconds is [tex]\(-2 \, \text{N}\)[/tex].