In a tennis serve, a .074 kg ball can be accelerated from rest to 40 m/s over a distance of 0.79 m.Find the magnitude of the average force exerted by the racket on the ball during the serve.
V ^2= 2ad =(40)^2,
2 * a * 0.79 = 1600,
1.58a = 1600,
a = 1600 / 1.58 = 1012.7 m/s^2.
F = ma = 0.74kg*1012.7 m/s^2 = 749.4N.
To find the magnitude of the average force exerted by the racket on the ball during the serve, we can use Newton's second law of motion:
F = m * a
Where:
F is the force exerted (what we want to find),
m is the mass of the ball (0.074 kg), and
a is the acceleration of the ball.
Since the ball starts from rest and accelerates to a final speed of 40 m/s over a distance of 0.79 m, we can use the following equation to find the acceleration:
v^2 = u^2 + 2*a*s
Where:
v is the final velocity (40 m/s),
u is the initial velocity (0 m/s, since the ball starts from rest),
a is the acceleration (what we want to find), and
s is the distance traveled (0.79 m).
Plugging in the values, we get:
(40 m/s)^2 = (0 m/s)^2 + 2 * a * 0.79 m
1600 m^2/s^2 = 2 * a * 0.79 m
Taking the square root and simplifying:
a = √(1600 m^2/s^2 / (2 * 0.79 m))
a ≈ 28.9 m/s^2
Now we can substitute the value of acceleration (28.9 m/s^2) into Newton's second law to find the force:
F = m * a
F = 0.074 kg * 28.9 m/s^2
F ≈ 2.14 N
Therefore, the magnitude of the average force exerted by the racket on the ball during the serve is approximately 2.14 Newtons.