Question

Very large forces are produced in joints when a person jumps from some height to the ground.

a) Calculate the force exerted if a 96kg person jumps from a 0.6m high ledge and lands stiffly, compressing joint material 3.4cm as a result. (Be certain to include the weight of the person.)

b) In practice the knees bend almost involuntarily to help extend the distance over which you stop. Calculate the force exerted if the stopping distance is 0.3m.

c) Determine the ratio of the force exerted with stiff legs to the weight of the person.

d) Determine the ratio of the force with bent legs to the weight of the person.

Answer 1

a)

m = mass of the person = 96 kg

h = height from which the person jumps = 0.6 m

vo = speed of the person just before landing = ?

Using conservation of energy

kinetic energy gained = gravitational potential energy lost

(0.5) m vo2 = mg h

vo2 = 2gh

vo2 = 2gh

vf = final speed = 0 m/s

d = stopping distance = 3.4 cm = 0.034 m

a = acceleration

Using the equation

vf2 = vo2 + 2 a d

02 = 2gh + 2 a d

0 = 2 (9.8) (0.6) + 2 a (0.034)

a = - 172.941 m/s2

force exerted is given as

F = ma

F = (96) (172.941 )

F = 1.66 x 104 N

b)

m = mass of the person = 96 kg

h = height from which the person jumps = 0.6 m

vo = speed of the person just before landing = ?

Using conservation of energy

kinetic energy gained = gravitational potential energy lost

(0.5) m vo2 = mg h

vo2 = 2gh

vo2 = 2gh

vf = final speed = 0 m/s

d = stopping distance = 0.3 m

a = acceleration

Using the equation

vf2 = vo2 + 2 a d

02 = 2gh + 2 a d

0 = 2 (9.8) (0.6) + 2 a (0.3)

a = - 19.6 m/s2

force exerted is given as

F = ma

F = (96) (19.6 )

F = 1881.6 N

c)

Ratio = F/mg = (1.66 x 104 )/((96) (9.8)) = 17.6

d)

Ratio = F/mg = (1.66 x 104 )/(1881.6) = 8.8