A 72 kg man drops to a concrete patio from a window only 0.60 m above the patio. He neglects to bend his knees on landing, taking 2.0 cm to stop.
(a) What is his average acceleration from when his feet first touch the patio to when he stops?
m/s2
(b) What is the magnitude of the average stopping force?
kN
I will be happy to critique your thinking.
a) First you calculate his speed of impact (the speed when he hits the ground).
For this you can use the formula:
v² = v0² + 2.a.x
where:
-v is your final speed
-v0 is your initial speed (0 m/s in this case)
-a is your acceleration (the gravitational acceleration in this case, being 9.81 m/s²)
-x, the distance travelled (being 0.6 m in this case)
so v² = 2*9.81 m/s²*0.6 m = 11.77 m²/s²
Now, we can use the same formula to calculate his average acceleration when he is slowing down.
v² = v0² + 2*a*x
- v being 0 m/s in this case
- v0² being 11.77 m²/s² (since he starts slowing down at this speed)
- x, being 0.02m
=> a = (v² - v0²)/(2*x) = (-11.77 m²/s²)/(2*0.02m) = - 294,25 m/s²
b) Since F = m*a
The average stopping force can be calculated as:
F = 72 kg* -294.25 m/s² = -21.186 kN
The negative sign for both the acceleration and stopping force means they are poiting upward, since we took the downward direction to be positive.
To calculate the average acceleration and the magnitude of the average stopping force, we need to use the laws of motion. Let's break down the problem into steps:
Step 1: Calculate the initial velocity (v₀) when the man hits the ground.
To find the initial velocity, we can use the equation of motion:
v² = v₀² + 2aΔd
Since the man is dropping from rest, v₀=0, and the equation simplifies to:
v² = 2aΔd
Rearranging the equation and plugging in the given values:
a = v² / (2Δd)
a = (0 - 2(0.6)) / (2(0.02))
a = -0.6 / 0.04
a = -15 m/s²
Note: The negative sign indicates that the acceleration is in the opposite direction to the motion.
Step 2: Calculate the average acceleration (a_avg) from the moment his feet first touch the patio to when he stops.
Since the acceleration is constant, the average acceleration is equal to the acceleration at any point during this time interval. Therefore, the average acceleration is -15 m/s².
(a) The average acceleration from when his feet first touch the patio to when he stops is -15 m/s².
Step 3: Calculate the magnitude of the average stopping force (F_avg).
To find the magnitude of the average stopping force, we can use Newton's second law of motion:
F_avg = m * a_avg
Plugging in the given values:
F_avg = 72 kg * (-15 m/s²)
F_avg = -1080 N (since force is a vector, we don't consider the negative sign here)
Step 4: Convert the average stopping force from Newtons to kilonewtons.
1 kN = 1000 N, so:
F_avg = -1080 N / 1000
F_avg = -1.08 kN
(b) The magnitude of the average stopping force is 1.08 kN (rounded to two decimal places).