Question

A runner whose mass is 54 kg accelerates from a stop to a speed of 7 m/s in 3 seconds. (A good sprinter can run 100 meters in about 10 seconds, with an average speed of 10 m/s.) (a) What is the average horizontal component of the force that the ground exerts on the runner's shoes? (b) How much work is done on the point-particle system by this force?

Answer 1

Answer:

a. $F=126N$

b. $E_K=1323J$

Explanation:

Given:

$m=54kg$

$v=7 m/s$

$t= 3s$

The runner force average to find given the equations

a.

$F=m*a$

$a=\frac{v}{t}$

$F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}$

$F=126N$

b.

Work done by the system by this force so

$W=F*d$

$W=E_K$

$E_K=\frac{1}{2}*m*v^2$

$E_K=\frac{1}{2}*54kg*(7m/s)^2$

$E_K=1323J$

Answer 2

(a) The average horizontal component of the force that the ground exerts on the runner's shoes is 126 N.

(b) The work done on the point-particle system by this force is 1,323 J.

Net force on the runner

The net force on the runner is determined from Newton's second law of motion as shown below.

Ff = ma

The average horizontal component of the force that the ground exerts on the runner's shoes is calculated as follows;

Ff = m(v/t)

Ff = 54 x (7/3)

Ff = 126 N

Work done by the force of friction

The work done by frictional force is calculated as follows;

W = ¹/₂mv²

W = 0.5 x 54 x 7²

W = 1,323 J

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