an 88.kg fireman slides 5.9 meters down a fire pole. He holds the pole, which exerts a 520N steady resistive force on the fireman. At the bottom he slows down to a stop in 0.43m by bending his knees. determine the acceleration while stopping and the time it takes for the fireman to stop after reaching the ground. I need help figuring out how to sole this.
his energy at the bottom is the change in potential energy, minus the work of friction
... m * g * h - 520 * h
... E = (88 kg * 9.8 m/s^2 - 520 N) * 5.9 m
his velocity is (from K.E. equation)
... v = √(2 * E / m)
while he is slowing down, his average velocity is
... (v + 0) / 2 = v / 2
his stopping time is
... t = .43 m / (v / 2)
his deceleration is
... a = v / t
To solve this problem, we can use Newton's second law of motion and the equations of motion.
First, let's determine the acceleration while stopping. Newton's second law states that the net force acting on an object is equal to its mass multiplied by its acceleration (F = m * a).
In this case, the force acting on the fireman is the resistive force exerted by the pole, which is 520 N. We can set this equal to the product of the fireman's mass and acceleration.
520 N = 88 kg * a
Rearranging the equation, we find:
a = 520 N / 88 kg
a = 5.91 m/s²
So, the acceleration while stopping is 5.91 m/s².
Next, let's determine the time it takes for the fireman to stop after reaching the ground.
We can use the equation of motion: v² = u² + 2as, where:
v = final velocity (0 m/s, since he comes to a stop)
u = initial velocity (velocity when the fireman reaches the ground)
a = acceleration (-5.91 m/s², as it acts opposite to the direction of motion)
s = displacement (0.43 m)
Since the acceleration acts opposite to the direction of motion, the initial velocity will be positive. We can consider the initial velocity to be the velocity of the fireman when he reaches the ground.
0² = u² + 2 * -5.91 m/s² * 0.43 m
Rearranging the equation, we find:
u² = 2 * 5.91 m/s² * 0.43 m
u² ≈ 5.06 m²/s²
Taking the square root of both sides, we find:
u ≈ √5.06 m/s
u ≈ 2.25 m/s
Now, we can use the equation of motion: v = u + at, where:
v = final velocity (0 m/s)
u = initial velocity (2.25 m/s)
a = acceleration (-5.91 m/s²)
t = time
0 = 2.25 m/s + (-5.91 m/s²) * t
Rearranging the equation, we find:
-2.25 m/s = -5.91 m/s² * t
Dividing both sides by -5.91 m/s², we find:
t = -2.25 m/s / -5.91 m/s²
t ≈ 0.38 s
So, the time it takes for the fireman to stop after reaching the ground is approximately 0.38 seconds.