A baseball player puts a baseball on a batting tee. The baseball bat is able to deliver 400,000 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 145g. What is the average acceleration?
To calculate the average acceleration, we need to use the formula:
Acceleration (a) = Change in velocity (Δv) / Time interval (Δt)
First, we need to calculate the change in velocity. Since we don't have the initial and final velocities directly, we can use the equation for work done:
Work (W) = Force (F) * Distance (d) = ΔKE (change in kinetic energy)
Here, the work done by the bat is equal to the change in kinetic energy of the ball:
W = ΔKE = 1/2 * m * (vf^2 - vi^2)
Where:
W = work done
m = mass of the ball (145g or 0.145kg)
vf = final velocity of the ball (which we need to find)
vi = initial velocity of the ball (assumed to be 0 m/s as it started from rest)
Given the power of the baseball bat as 400,000 W and contact time of 0.70 ms, we can calculate the work done. Power can be expressed as work done divided by time:
P = W / t
Rearranging the equation, we have:
W = P * t
Substituting the given values, we can find the work done by the bat:
W = (400,000 W) * (0.70 ms/1000) = 280 J
Now, let's find the final velocity using the equation for work done:
280 J = 1/2 * (0.145 kg) * (vf^2 - 0^2)
Simplifying the equation:
2 * 280 J = 0.145 kg * vf^2
vf^2 = (2 * 280 J) / 0.145 kg
vf^2 = 3862.069
Taking the square root of both sides:
vf = √3862.069 = 62.16 m/s
Now that we have the final velocity, we can calculate the average acceleration using the equation:
a = Δv / Δt
Given that the initial velocity (vi) is 0 m/s and Δt is 0.70 ms (or 0.0007 s), the average acceleration is:
a = (62.16 m/s - 0 m/s) / 0.0007 s
a = 88,800 m/s^2
Therefore, the average acceleration of the baseball is 88,800 m/s^2.