A 0.180 kg baseball, traveling 32.0 m/s, strikes the catcher's mitt, which recoils 13.0 cm in bringing the ball to rest. What was the average force (in newtons) applied by the ball to the mitt?
Kinetic energy loss equals work done pushing the mitt. Therefore,
Force = MV^2/(2X)
I was just wondering if you just square v and also what do you put in for x?
yes, V gets squared. X = 0.13 m
The force will be in Newtons.
To find the average force applied by the ball to the mitt, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = ma).
First, we need to determine the acceleration of the ball when it comes to rest. We can use the equation of motion:
(vf^2 - vi^2) = 2ad
Where vf is the final velocity (0 m/s since the ball comes to rest), vi is the initial velocity (32.0 m/s), a is the acceleration, and d is the displacement (13.0 cm or 0.13 m).
Plugging in the known values into the equation, we have:
(0 - 32.0^2) = 2a(0.13)
Simplifying further:
-1024.0 = 0.26a
Next, we need to solve for the acceleration (a):
a = -1024.0 / 0.26
a ≈ -3,938.46 m/s^2
Notice that the acceleration is negative, indicating a deceleration or slowing down of the ball.
Now that we have the acceleration, we can find the force by multiplying it by the mass of the ball:
F = ma
F = 0.180 kg × (-3,938.46 m/s^2)
F ≈ -708.5 N
The average force applied by the ball to the mitt is approximately 708.5 Newtons. The negative sign indicates that the force is in the opposite direction of the ball's initial motion.