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?
(b) What is the magnitude of the average stopping force?
physics - bobpursley, Thursday, December 11, 2008 at 8:11pm
I will be happy to critique your thinking.
physics - Christiaan, Thursday, December 11, 2008 at 8:14pm
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
-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
physics - Christiaan, Thursday, December 11, 2008 at 8:17pm
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.