A baseball pitcher can throw a baseball at a speed of 42 m/s. If the mass of the ball is 0.12 kg and the pitcher has it in his hand over a distance of 1.9 m, what is the average force exerted by the pitcher on the ball?
i already got the answer.
W = F d = (1/2) m v^2
F (1.9) = (1/2) (.12)(42^2)
f=.5x.12x42^2/1.9
To find the average force exerted by the pitcher on the ball, we can use Newton's second law of motion, which states that force is equal to the rate of change of momentum. The momentum of an object is calculated by multiplying its mass by its velocity.
First, let's calculate the initial momentum of the ball. We know the mass of the ball is 0.12 kg and the velocity at which it is thrown is 42 m/s. Therefore, the initial momentum is given by:
Initial momentum = mass × velocity
Initial momentum = 0.12 kg × 42 m/s
Next, let's calculate the final momentum of the ball. Since the ball is thrown at a constant velocity, its final momentum is the same as the initial momentum.
Final momentum = initial momentum = 0.12 kg × 42 m/s
Now, let's calculate the change in momentum of the ball:
Change in momentum = Final momentum - Initial momentum
Change in momentum = (0.12 kg × 42 m/s) - (0.12 kg × 0 m/s)
Change in momentum = 0.12 kg × 42 m/s
Finally, let's calculate the average force exerted by the pitcher on the ball using Newton's second law:
Average force = change in momentum / time
However, we need to calculate the time taken by the pitcher to exert this force. We can do this by dividing the distance the pitcher moves his hand by the velocity of the ball.
Time = distance / velocity
Time = 1.9 m / 42 m/s
Now that we have the time, we can calculate the average force:
Average force = (0.12 kg × 42 m/s) / (1.9 m / 42 m/s)
Average force = 5.04 N
Therefore, the average force exerted by the pitcher on the ball is 5.04 Newtons.