When a 58g tennis ball is served, it accelerates from rest to a speed of 45m/s. The impact with the racket gives the ball a constant acceleration over a distance of 44cm. What is the magnitude of the net force acting on the ball?
Vf^2 = Vo^2 + 2a*d.
a = (Vf^2-Vo^2)/2d,
a = ((45)^2-0) / 0.88 = 2301 m/s^2.
Fn = ma = 0.058kg * 2301 = 133.5 N.
NOTE:
44 cm = 0.44 m.
58 g. = 0.058 kg.
133.45
2 significant figures is 130N
When a 58-g tennis ball is served, it accelerates from rest to a speed of 45 m/s. The
impact with the racket gives the ball a constant acceleration over a distance of 44 cm.
What is the magnitude of the net force acting on the ball?
Well, that tennis ball must have had quite the racket! But let's get to the physics of it. We can use the second equation of motion: v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
First, let's convert the distance from centimeters to meters: 44 cm = 0.44 m.
Given:
Initial velocity (u) = 0 m/s (since it starts from rest)
Final velocity (v) = 45 m/s
Distance (s) = 0.44 m
Using the equation: v² = u² + 2as, we can rearrange it to solve for acceleration (a):
a = (v² - u²) / (2s)
Plugging in the values, we get:
a = (45² - 0²) / (2 * 0.44)
Calculating that, we find that the acceleration (a) is approximately 2314.77 m/s².
Now, let's calculate the net force (F) using the second law of motion: F = ma, where m is the mass of the tennis ball.
Given:
Mass (m) = 58 g = 0.058 kg
Plugging in the values, we get:
F = (0.058 kg) * (2314.77 m/s²)
Calculating that, we find that the magnitude of the net force acting on the ball is approximately 134.1533 N.
So, we can say that the net force acting on the ball is approximately 134.1533 N. Who knew tennis balls had such strong forces?
To determine the magnitude of the net force acting on the ball, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Given:
Mass of the tennis ball (m) = 58g = 0.058kg (since 1 g = 0.001 kg)
Final velocity of the ball (v) = 45 m/s
Distance over which the ball is accelerated (s) = 44 cm = 0.44 m
First, we need to find the acceleration (a) of the ball using the kinematic equation for uniformly accelerated motion:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity (0 m/s since the ball starts from rest)
a = acceleration
s = distance
Rearranging the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
a = (45^2 - 0^2) / (2 * 0.44)
Calculating the acceleration:
a = 1822.5 m/s^2
Now that we have the acceleration, we can calculate the magnitude of the net force (F) acting on the ball using Newton's second law:
F = m * a
F = 0.058 kg * 1822.5 m/s^2
Calculating the net force:
F ≈ 105.8 N
Therefore, the magnitude of the net force acting on the ball is approximately 105.8 Newtons.