A person jumps from a tower 5 m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70m. If the mass of his torso (excluding legs) is 50kg, find the force exerted on his torso by his legs during the deceleration.
To find the force exerted on the person's torso by his legs during deceleration, we need to calculate the deceleration first.
Given:
Initial velocity, u = 0 (since the person jumps from rest)
Final velocity, v = ?
Distance, s = 0.70 m
Acceleration, a = ?
Mass of the torso, m = 50 kg
We can use the equation of motion:
v^2 = u^2 + 2as
Rearranging the equation to solve for final velocity 'v':
v^2 = 2as
v = √(2as)
v = √(2 * 0.70 * 9.8) [Acceleration due to gravity, g ≈ 9.8 m/s^2]
v ≈ 4.116 m/s
The final velocity of the person's torso just before coming to rest is approximately 4.116 m/s.
To calculate the deceleration:
Deceleration, a = (v - u) / t
where t is the time taken to decelerate.
Since the distance traveled while decelerating is not given, we can use the equation of motion:
s = u * t + (1/2) * a * t^2
0.70 = (0 * t) + (1/2 * a * t^2)
0.70 = (1/2 * a * t^2)
Rearranging the equation to solve for deceleration 'a':
a = (2 * s) / t^2
Applying the values, we have:
a = (2 * 0.70) / t^2
As the time taken to decelerate is not given, we cannot calculate the exact deceleration or the force exerted by the legs on the torso without knowing the time.
To find the force exerted on the torso by the legs during deceleration, we will need to use the conservation of energy and the formulas related to work and force.
First, let's find the initial potential energy when the person jumps from the tower. The potential energy (PE) is given by the formula:
PE = m * g * h
Where:
m = mass of the torso = 50 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the tower = 5 m
Substituting the values, we get:
PE = 50 kg * 9.8 m/s^2 * 5 m
PE = 2450 Joules
Next, let's find the final potential energy when the person lands on the ground and his torso decelerates. The final potential energy is zero since the person is at ground level. So we can equate the initial potential energy to the work done on the person's torso during deceleration.
PE = Work
Work = Force * Distance
The distance over which the force is applied is given as 0.70 m.
2450 Joules = Force * 0.70 m
Now, we can rearrange the equation to solve for Force:
Force = 2450 Joules / 0.70 m
Force = 3500 Newtons
Therefore, the force exerted on the person's torso by their legs during deceleration is approximately 3500 Newtons.