Question

A 97-kg sprinter wishes to accelerate from rest to a speed of 11 m/s in a distance of 21 m. What coefficient of static friction is required between the sprinter's shoes and the track?

Answer 1

coefficient of static friction required between the sprinter's shoes and the track is  u=0.293.

Explanation:-  

The frictional coefficient is found from newton’s second law

$\Sigma F=m u g \ldots \ldots \ldotseqn(1)$

We know that  

$F=m a \ldots \ldots \ldots{eqn}(2)$

Compare the Eqn (1) and eqn (2)  

$m u g=m a$

Mass (m) cancel each other both side

So

$u g=a$

Coefficient of static friction    

$u=\frac{a}{g} \ldots \ldots {eqn}(3)$

We know that  

Kinetics  $V^{2}=V_{0}^{2}+2 .a .\Delta s$  

$V=r e s t \text { of speed }=11 \mathrm{m} / \mathrm{s}$

$V_{0}=\text { initial speed}=0$

$\Delta s=d i s t a n c e=21 m$

$a=\text { acceleration }=\text { unknown }$

$a=\frac{\left(V^{2}-V_{0}^{2}\right)}{2 \Delta s}$

$a=\frac{\left(11^{2}-0\right)}{(2 \times 21)}$

$a=\frac{121}{42}$

$a=2.88 \mathrm{m} / \mathrm{s}^{2}$

Substitute the value of acceleration in eqn (3)

So that

Coefficient of static friction.

$u=\frac{\left(2.88 m / s^{2}\right)}{\left(9.8 m / s^{2}\right)}$

$g=9.8 \text { is earth gravity }$  

Coefficient of static friction     $u=0.293.$