A baseball player puts a baseball on a batting tee. The baseball bat is able to deliver 4.0x10^5 W of power and is in contact with the ball for 0.70 ms and a distance of 1.4 cm. The mass of the ball is 145 g. What is the next force applied to the ball, and what is its average acceleration?
Incorrect attempt:
P = de/dt
400000 = F_net * 0.014 / 0.0007
...
F_net = 20000 N
Answer should be 200000.
I think you did it right. My teacher got the same answer as u. Why would the answer be 200000?
Well, it seems like the baseball player forgot to sign a contract with their physics teacher for a better answer. Let's use our funny physics powers to solve this!
To find the force applied to the ball, we'll start with the power equation P = F * v, where P is power, F is force, and v is velocity. But wait, we don't have the velocity! Don't panic, we can relate velocity, time, and distance using the equation v = d/t.
Given that the ball is in contact with the bat for 0.70 ms and travels a distance of 1.4 cm, let's calculate the velocity using v = d/t:
v = 0.014 m / (0.70 x 10^-3 s) # Convert to meters and seconds
v ≈ 20 m/s # Approximately
Now, let's rearrange the power equation to solve for force:
P = F * v
F = P / v
F = 4.0 x 10^5 W / (20 m/s)
F = 2.0 x 10^4 N
Oops, looks like your attempt was correct! The force applied to the ball is indeed 20,000 N, not 200,000 N as you mentioned. As for the average acceleration, we'll use the equation F = ma, where m is the mass and a is the acceleration.
F = ma
a = F / m
a = (20,000 N) / (0.145 kg) # Convert mass to kg
a ≈ 137,931 m/s² # Approximately
So, the average acceleration is approximately 137,931 m/s². Hope this explanation made you crack a smile while learning physics!
To find the correct answer, we can start by using the definition of power:
Power (P) = work done (W) / time (t)
In this case, the work done is the force applied (F_net) multiplied by the distance the ball is in contact with the bat (d). Rearranging the formula, we have:
P = (F_net * d) / t
We can then solve for the force applied (F_net):
F_net = (P * t) / d
Now, substituting the given values into the formula:
P = 4.0x10^5 W
t = 0.70 ms = 0.70 * 10^-3 s
d = 1.4 cm = 1.4 * 10^-2 m
F_net = (4.0x10^5 W * 0.70 * 10^-3 s) / (1.4 * 10^-2 m)
Simplifying the expression:
F_net = (2.8x10^2 N * 0.70 * 10^-3 s) / (1.4 * 10^-2 m)
F_net = 200 N
Therefore, the force applied to the ball is 200 N, not 20000 N as in your incorrect attempt.
I have come to believe the answer given was wrong after seeing so many other errors in the answers.
Why would the power not be wasted in heat and compression of the ball to a great extent?
In .7*10^-3 seconds the bat accelerates the ball from zero to final speed over a distance of .014 meters
I assume the acceleration was constant during that time
so the average speed over that distance was .014 m/.0007 s = 20 m/s
thus the final speed is 40 m/s
force = change in momentum/time
= mass (change in velocity/change in time)
change in velocity/change in time
= average acceleration = 40/.0007
= 57142 m/s^2
F = m a = .145 * 57142 = 8286 N