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.
3 answers
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
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
I think you did it right. My teacher got the same answer as u. Why would the answer be 200000?
0 answers