A CRICKET BALL OF MASS 0.4kg MOVING WITH A VELOCITY OF 72km/h IS BROUGHT TO REST
BY A PLAYER . WHAT IS THE MAGNITUDE OF THE AVERAGE FORCE EXERTED BY THE PLAYER?
To find the magnitude of the average force exerted by the player to bring the cricket ball to rest, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a).
First, let's convert the velocity of the cricket ball from km/h to m/s. Since 1 km/h equals 1000 m/3600 s, the velocity is:
72 km/h = (72 * 1000) / 3600 = 20 m/s.
Next, we need to determine the acceleration of the cricket ball. Since the ball is brought to rest, its final velocity (vf) is 0 m/s. Using the equation vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the distance, we can solve for acceleration.
Given that the initial velocity (vi) = 20 m/s and final velocity (vf) = 0 m/s, we can rearrange the equation to solve for acceleration (a):
0 = (20)^2 + 2a * d.
Since we are not given the distance, we can assume that the cricket ball comes to rest in a short distance, making the value of 'd' negligible. Therefore, we can simplify the equation to:
0 = 400 + 2a * 0,
0 = 400.
Since the distance is negligible, the acceleration (a) is 0. Therefore, the average force exerted by the player is also 0.
Hence, the magnitude of the average force exerted by the player to bring the cricket ball to rest is 0.
depends on how long it took to stop the ball, or how far it traveled while being stopped.
Surely you can see that a gentle push (small force) exerted over many meters would do the job, as would a strong force imparted in an instant.