A 71.1 kg man steps out a window and falls (from rest) 1.24 m to a sidewalk. What is his speed just before his feet strike the pavement?
If the man falls with his knees and ankles locked, the only cushion for his fall is a 0.500 cm give in the pads of his feet. Calculate the average force exerted on him by the ground in this situation. This average force is sufficient to cause cartilage damage in the joints or to break bones.
To solve the first part of the question and find the speed of the man just before his feet strike the pavement, we can use the principle of conservation of energy.
1. First, let's calculate the potential energy the man has at the initial height (the window) using the formula:
Potential Energy = mass * gravitational acceleration * height
Given:
Mass (m) = 71.1 kg
Gravitational acceleration (g) = 9.8 m/s^2
Height (h) = 1.24 m
Potential Energy (PE) = 71.1 kg * 9.8 m/s^2 * 1.24 m
2. Next, let's calculate the kinetic energy the man has just before his feet strike the pavement. Since the man falls from rest, all the potential energy he had is converted into kinetic energy.
Kinetic Energy = Potential Energy
Kinetic Energy (KE) = PE
3. Now, we can find the speed of the man just before his feet strike the pavement using the formula:
Kinetic Energy = 0.5 * mass * velocity^2
Rearranging the formula:
velocity^2 = (2 * Kinetic Energy) / mass
velocity = sqrt((2 * KE) / mass)
Substituting the known values:
velocity = sqrt((2 * PE) / mass)
Finally, we can calculate the speed:
velocity = sqrt((2 * (71.1 kg * 9.8 m/s^2 * 1.24 m)) / 71.1 kg)
Solving this equation will give us the speed of the man just before his feet strike the pavement.
Moving on to the second part of the question, to calculate the average force exerted on the man by the ground, we can use the concept of impulse and momentum.
1. First, let's calculate the change in momentum of the man when he hits the ground. Since he falls with his knees and ankles locked, his velocity will change from the initial velocity (calculated in the first part) to zero.
Change in momentum (Δp) = mass * (final velocity - initial velocity)
Since final velocity is 0, it simplifies to:
Δp = -mass * initial velocity
2. The average force exerted on the man can be calculated using the formula:
Impulse = Average force * time
Impulse = Δp (change in momentum)
Rearranging the formula:
Average force = Δp / time
We need to determine the time over which the impulse is applied. For this, we will consider the 0.500 cm give in the pads of his feet. Assuming this give is the time it takes for the man's feet to come to rest, we can use this value as the time.
Convert the give (0.500 cm) to meters (0.005 m).
Finally, we can calculate the average force:
Average force = Δp / time
Substituting the known values, we can solve for the average force exerted on the man by the ground.
Remember to substitute the calculated values and solve the equations step by step to find the solutions to both parts of the question.
To find the man's speed just before his feet strike the pavement, we can use the equation for gravitational potential energy:
Potential energy (PE) = mass (m) * acceleration due to gravity (g) * height (h)
PE = m * g * h
where:
m = 71.1 kg (mass of the man)
g = 9.8 m/s^2 (acceleration due to gravity)
h = 1.24 m (height fallen)
PE = 71.1 kg * 9.8 m/s^2 * 1.24 m
PE = 870.2472 J
Next, we can equate the potential energy to the kinetic energy just before the feet strike the pavement:
Potential energy (PE) = Kinetic energy (KE)
KE = 870.2472 J
The equation for kinetic energy is:
Kinetic energy (KE) = 1/2 * mass * velocity^2
1/2 * m * v^2 = 870.2472 J
Rearranging the equation, we get:
v^2 = (2 * 870.2472 J) / 71.1 kg
v^2 = 24.5002 m^2/s^2
Taking the square root of both sides, we find:
v = sqrt(24.5002 m^2/s^2)
v = 4.949 m/s
Therefore, the man's speed just before his feet strike the pavement is approximately 4.949 m/s.
To calculate the average force exerted on the man by the ground, we can use the work-energy principle. The work done by the ground is equal to the change in kinetic energy during impact.
The work done by the ground (W) = change in kinetic energy (ΔKE)
ΔKE = KE_final - KE_initial
As the man comes to a stop during impact, his final kinetic energy (KE_final) is 0 J. Therefore:
ΔKE = -KE_initial
Since the initial kinetic energy (KE_initial) is given by:
KE_initial = 1/2 * mass * velocity^2
KE_initial = 1/2 * 71.1 kg * (4.949 m/s)^2
KE_initial = 885.468 J
So, the change in kinetic energy is:
ΔKE = -885.468 J
Since work done is equal to the force applied multiplied by the displacement:
W = F * d
where:
W = -885.468 J (negative sign represents work done against the motion)
d = 0.500 cm = 0.005 m (displacement)
We can rearrange the equation to solve for the average force:
F = W / d
F = -885.468 J / 0.005 m
F ≈ -177,093.6 N
Therefore, the average force exerted on the man by the ground is approximately -177,093.6 N (negative sign indicates that the force is directed upwards, opposing the motion).
Total energy during the fall is constant.
Thus loos of potential energy = gain in kinetic energy
m g h = (1/2) m v^2
so
v = sqrt (2 gh)
Force * distance = energy
Here the energy is m g h, the loss of potential energy during the fall.