Question

Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet and legs.

(a) Calculate the magnitude of the force needed to stop the downward motion of a jogger’s leg, if his leg has a mass of 13.0 kg, a speed of 6.00 m/s, and stops in a distance of 1.50 cm. (Be certain to include the weight of the 75.0-kg jogger’s body.)
(b) Compare this force with the weight of the jogger. (74)[OS 54] (a)

Answer 1

Part a)

`F = 15600 N`

Part b)

force is 21.2 times more than the weight of the person

Step-by-step explanation:

Part a)

As it is given that the distance after which he stopped is given as

d = 1.50 cm

here finally it stops so final speed is given as

`v_f = 0`

initial speed is given as

`v_i = 6 m/s`

now by the equation of kinematics we know that

`v_f^2 - v_i^2 = 2a d`

`0^2 - 6^2 = 2(a)(0.015)`

`a = -1200 m/s^2`

Now the force on the leg is given as

`F = ma`

m = mass of leg = 13 kg

`F = (13 kg)(-1200 m/s^2)`

`F = 15600 N`

Part b)

Force due to weight of the object

`F_g = Mg`

here we know

M = 75.0 kg

`F_g = 75(9.81)`

`F_g = 735.75 N`

Now we know that the ratio of the weight with the force on leg is given as

`(F)/(F_g) = (15600)/(735.75) = 21.2 N`

so force is 21.2 times more than the weight of the person