A person jumps from the roof of a house 4.0-m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.66m. If the mass of his torso (excluding legs) is 42kg , what is the magnitude of the average force exerted on his torso by his legs during deceleration?
To find the magnitude of the average force exerted on the person's torso by his legs during deceleration, we need to use the principle of conservation of energy.
The potential energy before the jump is given by the formula PE = mgh, where m is the mass of the person's torso, g is the acceleration due to gravity, and h is the height of the jump.
The initial potential energy is:
PE_initial = mgh = 42 kg * 9.8 m/s^2 * 4.0 m = 1646.4 J
During the deceleration, the potential energy is converted into work done by the force exerted by the legs. The work formula is given by W = F * d, where W is the work done, F is the force applied, and d is the distance over which the force is applied.
The work done is equal to the change in potential energy, so we can write:
W = ΔPE = PE_final - PE_initial
Since the person bends his knees and his torso decelerates over a distance of 0.66m, the final potential energy is:
PE_final = mgh_final = 42 kg * 9.8 m/s^2 * 0.66 m = 267.528 J
Substituting the values into the work formula, we have:
267.528 J = F * 0.66 m
To find the force exerted by the legs, we can rearrange the equation:
F = 267.528 J / 0.66 m ≈ 405.27 N
Therefore, the magnitude of the average force exerted on the person's torso by his legs during deceleration is approximately 405.27 Newtons.