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
To answer these questions, we can use the principles of kinematics and Newton's laws of motion.
(a) To find the average acceleration, we need to determine the change in velocity and the time it takes for the man to stop. First, let's find the initial velocity of the man just as he touches the patio. We can do this using the equation for free fall:
v^2 = u^2 + 2as
Where:
- v is the final velocity (in this case, 0 m/s because he stops)
- u is the initial velocity (unknown)
- a is the acceleration due to gravity (approximately 9.8 m/s^2)
- s is the distance the man travels during free fall (0.60 m)
Rearranging the equation, we get:
u = sqrt(v^2 - 2as)
Plugging in the values, we have:
u = sqrt(0 - 2 * 9.8 * 0.60)
u = sqrt(-11.76)
(Note: The negative sign indicates that the initial velocity is directed downward.)
Since we are interested in the magnitude of the velocity, we will take the positive square root:
u = 3.42 m/s (approximately)
Now, we can calculate the time it takes for the man to stop. We'll use the following equation of motion:
v = u + at
Where:
- v is the final velocity (0 m/s)
- u is the initial velocity (-3.42 m/s)
- a is the average acceleration (unknown, what we're trying to find)
- t is the time it takes to stop (unknown)
Rearranging the equation, we get:
t = (v - u) / a
Plugging in the values, we have:
0 = (-3.42) + a * t
Given that he takes 2.0 cm to stop (or 0.02 m), we can write:
0.02 = (-3.42) * t + (1/2) * a * t^2
We have two unknowns in this equation, t and a. However, we can solve them by using the initial equation, v = u + at, and substituting zero for v and rearranging to solve for t:
t = -u / a
Substituting the value of u = -3.42 m/s, we get:
0.02 = (-3.42) * (-3.42) / a
0.02 = 11.6764 / a
Solving for a, we have:
a = 11.6764 / 0.02 = 583.82 m/s^2
Therefore, the average acceleration from when his feet first touch the patio to when he stops is 583.82 m/s^2.
(b) To find the magnitude of the average stopping force, we can use Newton's second law of motion:
F = ma
Where:
- F is the force (what we're trying to find)
- m is the mass of the man (72 kg)
- a is the average acceleration (583.82 m/s^2)
Plugging in the values, we have:
F = 72 kg * 583.82 m/s^2 = 42099.84 N
Since 1 kilonewton (kN) is equal to 1000 N, we can convert the force to kilonewtons:
F = 42099.84 N / 1000 = 42.1 kN
Therefore, the magnitude of the average stopping force is 42.1 kN.
vf^2=vi^2+2ad solve for a.
force=mass*a