When a 51.7 gram tennis ball is served, it accelerates from rest to a speed of 41.2 m/s. The impact with the racket gives the ball a constant acceleration over a distance of 49.5 cm. What is the magnitude of the net force acting on the ball in newtons?

F = m*a

We first need to compute the acceleration, a.

V = sqrt(2aX)
41.2 = sqrt(2*a*0.495)
1697 = 2*a*0.495
a = 1714 m/s^2

F = m*a
= 0.0157 kg*1714 m/s^2
= 88.6 N

Another way to solve this is to equate the final kinetic energy to F*X (work done)

F = (1/2)MV^2/X
= (0.5)(0.0517)*41.2^2/(0.495)
= 88.6 N

thank you so much

To find 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 product of its mass and acceleration. Mathematically, it can be expressed as:

F_net = m * a

In this case, the mass of the tennis ball is given as 51.7 grams, which we need to convert to kilograms since the SI unit for mass is kilogram. We can do this by dividing the mass by 1000:

m = 51.7 grams ÷ 1000 = 0.0517 kg

The acceleration of the ball can be determined from the given information about its initial and final speeds, as well as the distance over which it accelerates. We'll use the following kinematic equation to calculate the acceleration:

v_f^2 = v_i^2 + 2 * a * d

Where:
v_f = final velocity = 41.2 m/s
v_i = initial velocity = 0 m/s (since it starts from rest)
a = acceleration (which is constant)
d = distance covered = 49.5 cm = 0.495 m (converted to meters)

Rearranging the equation to solve for acceleration (a):

a = (v_f^2 - v_i^2) / (2 * d)

Plugging in the values:

a = (41.2^2 - 0^2) / (2 * 0.495)
a = 1694.44 m^2/s^2 / 0.99
a ≈ 1710.606 m/s^2

Now, we have the mass (m = 0.0517 kg) and acceleration (a ≈ 1710.606 m/s^2). We can substitute these values into Newton's second law equation to find the net force:

F_net = m * a
F_net = 0.0517 kg * 1710.606 m/s^2
F_net ≈ 88.481 N

Therefore, the magnitude of the net force acting on the tennis ball is approximately 88.481 newtons.