Question

The tibia bone in the lower leg of an adult human will break if the compressive force on it exceeds about 4×105N (we assume that the ankle is pushing up). Suppose that someone step off a chair that is 0.40 m above the floor. If landing stiff-legged on the surface below, what minimum stopping distance does he need to avoid breaking his tibias? Assume that the mass of the person is 64 kg

Answer 1

Distance = 6.27 x 10⁻⁴m

Step-by-step explanation:

The speed just before hitting the ground may be expressed as

v = √2gh, where g is the acceleration due to gravity and h is the height

v = √(2 x 9.81 x 0.4) = 2.8 m/s

Using the work -energy theorem, we know that any work done on the object produces a change of kinetic energy in the object. Hence

Net Work = Final Kinetic Energy - Initial Kinetic Energy

The initial condition is stationary hence the initial Kinetic Energy = 0

Net Work may be written as

Net Work = Force x Distance = Final Kinetic Energy

4 x 10⁵ x d = (1/2) x mass x velocity²

d = {(1/2) x 64 x 2.8²}/(4 x 10⁵) = 6.27 x 10⁻⁴m

6.27 x 10⁻⁴m is the minimum stopping distance to avoid breaking his tibias.