A person jumps from the roof of a house 3.6 -m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 {\rm m}. If the mass of his torso (excluding legs) is 46 kg. Find the average force exerted on his torso by his legs during deceleration.

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To find the average force exerted on the person's torso by his legs during deceleration, we can use the concept of work.

1. First, let's determine the change in potential energy of the person's torso. The change in potential energy can be calculated using the formula:

ΔPE = mgΔh

where ΔPE is the change in potential energy, m is the mass of the torso, g is the acceleration due to gravity, and Δh is the change in height.

Given that the height is 3.6 m and the mass of the torso is 46 kg, the change in potential energy is:

ΔPE = (46 kg)(9.8 m/s^2)(-3.6 m) [Note: We take -3.6 m since the person is falling downwards]

2. Next, let's calculate the work done on the torso as it decelerates. The work done can be determined using the formula:

W = Fd

where W is the work done, F is the force, and d is the distance over which the force is applied.

Given that the distance is 0.70 m, we can rearrange the equation to solve for the force:

F = W/d

3. Finally, we can substitute the value of work (which is equal to the change in potential energy) into the equation to find the force:

F = ΔPE/d

Now, let's substitute the values and calculate the force:

ΔPE = (46 kg)(9.8 m/s^2)(-3.6 m) = -1608.48 J

F = (-1608.48 J) / (0.70 m)

F ≈ -2297.83 N

Therefore, the average force exerted on the person's torso by his legs during deceleration is approximately 2297.83 N.

To find the average force exerted on the torso by the legs during deceleration, we can use the concept of work and energy. We need to find the change in potential energy as the person falls from the roof and decelerates over a distance.

First, let's find the change in potential energy.

The change in potential energy is given by the formula:

ΔPE = mgh

Where:
ΔPE = Change in potential energy
m = Mass of the torso
g = Acceleration due to gravity (approximately 9.8 m/s^2)
h = Height of the roof

Substituting the given values:

ΔPE = (46 kg) * (9.8 m/s^2) * (3.6 m)

ΔPE = 1612.8 Joules

Now, let's find the work done by the legs during deceleration.

The work done is given by the formula:

Work = Force * Distance

In this case, the force is the average force exerted on the torso by the legs, and the distance is the approximate distance over which the torso decelerates.

So, we need to find the work done by the legs:

Work = Average Force * Distance

Substituting the given values:

Work = Average Force * 0.70 m

We know that the work done is equal to the change in potential energy:

Work = ΔPE

So, we can equate the two equations:

Average Force * 0.70 m = 1612.8 Joules

Now, we can solve for the average force exerted by the legs:

Average Force = 1612.8 Joules / 0.70 m

Average Force ≈ 2318.29 N

Therefore, the average force exerted on the torso by the legs during deceleration is approximately 2318.29 Newtons.