When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 75 kg man just before contact with the ground has a speed of 6.9 m/s.
(a) In a stiff-legged landing he comes to a halt in 2.0 ms. Find the average net force that acts on him during this time.
258750 N
(b) When he bends his knees, he comes to a halt in 0.10 s. Find the average force now.
5175 N
(c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of these forces, find the force of the ground on the man in parts (a) and (b).
stiff legged landing
258750 N
bent legged landing???
N
I figured out parts a and b. On part c I thought the same force would be exerted upward and downward, but that is not the case for the bent legged landing. Why? and how do you solve?
In the case of a stiff-legged landing, the man comes to a halt in a very short time (2.0 ms), thus experiencing a significant change in velocity. The average net force acting on him during this time is responsible for bringing him to a stop, and it is equal to his mass multiplied by the rate of change of velocity.
Using the formula F = ma, where F is the net force, m is the mass, and a is the acceleration, we can calculate the force in part (a):
F = m * a
Given:
mass, m = 75 kg
change in velocity, Δv = final velocity - initial velocity = 0 - 6.9 m/s = -6.9 m/s
time, t = 2.0 ms = 2.0 x 10^(-3) s
Acceleration, a = Δv / t = (-6.9 m/s) / (2.0 x 10^(-3) s) = -3450 m/s^2 (negative sign indicates deceleration)
Substituting the values back into the formula:
F = (75 kg) * (-3450 m/s^2) = -258,750 N
So, the average net force acting on the man in a stiff-legged landing is -258,750 N (negative sign indicates that it acts in the opposite direction of motion).
Now, let's move on to part (b) - the bent legged landing. In this case, it takes the man 0.10 s to come to a halt, which is a longer duration than in the stiff-legged landing. The force required to bring him to a stop in this time will be smaller. Again, we can use the formula F = ma to calculate the force.
Given:
mass, m = 75 kg
change in velocity, Δv = 0 - 6.9 m/s = -6.9 m/s
time, t = 0.10 s
Acceleration, a = Δv / t = (-6.9 m/s) / (0.10 s) = -69 m/s^2
Substituting the values back into the formula:
F = (75 kg) * (-69 m/s^2) = -5,175 N
So, the average net force acting on the man in a bent legged landing is -5,175 N (again, negative sign indicates that it acts in the opposite direction of motion).
In part (c), the question asks about the force of the ground on the man in both the stiff-legged and bent legged landings. Note that in both cases, the force of gravity always points downward with a magnitude of mg, where m is the mass and g is acceleration due to gravity.
However, in a stiff-legged landing, the net force exerted by the ground on the man must be equal to the negative of the average net force calculated in part (a) (-258,750 N). This upward force of the ground counteracts the downward force of gravity to bring the man to a halt.
So, in a stiff-legged landing, the force exerted by the ground on the man is +258,750 N (opposite in direction to the force of gravity).
On the other hand, in a bent legged landing, we already calculated the average net force to be -5,175 N. Since the man comes to a halt in 0.10 s, with a smaller force acting against his motion, the force exerted by the ground on him will also be smaller.
Therefore, in a bent legged landing, the force exerted by the ground on the man is in the direction of motion and has a magnitude of 5,175 N.
In a stiff-legged landing, the average net force acting on the man during deceleration is equal to the mass of the man multiplied by the change in velocity divided by the time taken to come to a halt. This gives us:
(a) Average net force = (Change in velocity) / (Time) = (0 - 6.9 m/s) / (2.0 ms)
Simplifying this, we get:
Average net force = -6.9 m/s / 2.0 ms = -3450 N
Since the average net force is negative, it means the force is acting in the opposite direction of the initial velocity. Therefore, the force of the ground on the man is upward.
Now let's consider the bent-legged landing. In this case, the time taken to come to a halt is longer (0.10 s), which allows for a smaller average net force. Using the same formula as before, we get:
(b) Average net force = (0 - 6.9 m/s) / (0.10 s) = -69 N
Again, the negative sign indicates that the force is acting opposite to the initial velocity, which means the force of the ground on the man is upward.
Therefore, it seems there is an error in the given answer for part (c) of the bent-legged landing. The force of the ground on the man should also be -69 N, indicating an upward force.
To summarize:
(a) Stiff-legged landing: Average net force = -3450 N (upward force)
(b) Bent-legged landing: Average net force = -69 N (upward force)