Question

A 54-kg ice skater pushes off the wall of the rink, giving herself an initial speed of 3.5 m/s . She then coasts with no further effort. The frictional coefficient between skates and ice is 0.024. How far does she go?

Answer 1

Answer:

Displacement is 72.917 m .

Explanation:

Given :

Mass of cyclist , m = 54 kg .

Initial speed , u = 3.5 m/s .

Frictional coefficient between skates and ice is , $\mu= 0.024$ .

We need to find its the displacement produced by it .

Frictional Force , $F=\mu (mg)=0.024\times 54\times 3.5=4.536\ N.$

Therefore, its acceleration , $a=\dfrac{F}{m}=\dfrac{4.536}{54}=0.084\ .$

We know , by equation of motion .

$v^2-u^2=2\times a \times s\\\\s=\dfrac{v^2-u^2}{2\times a}=\dfrac{0-3.5^2}{2 \times (-0.084)}=72.917\ m.$

Hence , this is the required solution .