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)

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aristocles aristocles · Answer 1 · 2019-08-09T10:06:16+02:00

Answer:

Part a)

[tex]F = 15600 N[/tex]

Part b)

force is 21.2 times more than the weight of the person

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

[tex]v_f = 0[/tex]

initial speed is given as

[tex]v_i = 6 m/s[/tex]

now by the equation of kinematics we know that

[tex]v_f^2 - v_i^2 = 2a d[/tex]

[tex]0^2 - 6^2 = 2(a)(0.015)[/tex]

[tex]a = -1200 m/s^2[/tex]

Now the force on the leg is given as

[tex]F = ma[/tex]

m = mass of leg = 13 kg

[tex]F = (13 kg)(-1200 m/s^2)[/tex]

[tex]F = 15600 N[/tex]

Part b)

Force due to weight of the object

[tex]F_g = Mg[/tex]

here we know

M = 75.0 kg

[tex]F_g = 75(9.81)[/tex]

[tex]F_g = 735.75 N[/tex]

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

[tex]\frac{F}{F_g} = \frac{15600}{735.75} = 21.2 N[/tex]

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