A physics teacher (mass = 84.5 kg) is jogging through the woods and runs straight into a large oak tree at 3.5 m/s. Rebound speed is measured at 2.3 m/s in the opposite direction. If the time of contact with the tree is 45 milliseconds, what is the magnitude of the force that the tree exerts on the teacher?
To find the magnitude of the force that the tree exerts on the teacher, you can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration.
First, let's calculate the change in velocity (Δv) of the teacher. The initial velocity (vi) is 3.5 m/s, and the final velocity (vf) is -2.3 m/s (since it's in the opposite direction). Therefore, the change in velocity is:
Δv = vf - vi
= -2.3 m/s - 3.5 m/s
= -5.8 m/s
Next, we need to convert the time of contact (t) to seconds. The given time is 45 milliseconds, so we divide it by 1000 to convert it to seconds:
t = 45 milliseconds / 1000
= 0.045 seconds
Now, we can calculate the acceleration (a) using the formula:
a = Δv / t
Substituting the values:
a = -5.8 m/s / 0.045 s
= -128.9 m/s²
Since the teacher is decelerating, the acceleration is negative.
Finally, we can find the magnitude of the force using Newton's second law:
force = mass x acceleration
= 84.5 kg x -128.9 m/s²
≈ -10872.05 N
Therefore, the magnitude of the force that the tree exerts on the teacher is approximately 10872.05 Newtons.