Answer:
The average horizontal force on the ball while it was in contact with the wall is -58.33N.
Explanation:
The average horizontal force can be found employing Newton's second law:
[tex]F = ma[/tex] (1)
Where F is the force, m is the mass and a is the acceleration.
The acceleration can be determined employing the equations for a Uniformly Accelerated Rectilinear Motion:
[tex]v_{f} = v_{i} + at[/tex] (2)
Where [tex]v_{f}[/tex] is the final velocity, [tex]v_{i}[/tex] is the initial velocity and t is the time.
Then a will be isolated from equation 2:
[tex]a = \frac{v_{f}-v_{i}}{t}[/tex] (3)
Replacing equation (3) in equation (1) it is gotten:
[tex]F = m\frac{v_{f}-v_{i}}{t}[/tex]
[tex]F = (0.35Kg)\frac{(-5.5m/s-7.0m/s)}{(0.075s)}[/tex]
[tex]F = (0.35Kg)\frac{(-12.5m/s)}{(0.075s)}[/tex]
[tex]F = -58.33Kg.m/s{2}[/tex]
But [tex]1N = 1Kg.m/s^{2}[/tex], therefore:
[tex]F = -58.33N[/tex]
The negative sign means that the force is acting in opposite direction to the movement of the ball (Newton's third law).
So, the average horizontal force on the ball while it was in contact with the wall is -58.33N.
Answer:
-58.33 N
Explanation:
From Newton's second law of motion,
F = m(v-u)/t....................... Equation 1
Where F = Average horizontal force, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t = time.
Given: m = 350 g = 0.35 kg, u = 7.0 m/s, v = -5.5 m/s, t = 0.075 s
Substitute into equation 1
F = 0.35(-5.5-7)/0.075
F = 0.35(-12.5)/0.075
F = -58.33 N.
Hence the average horizontal force on the ball while in contact with the wall = -58.33 N
