Question

In short-track speed skating, the track has straight sections and semicircles 16 m m/s .Part A What is the horizontal force on the skater? part B What is the ratio of this force to the skater's weight?kg skater goes around the turn at a constant 11m/s m in diameter. Assume that a 64kg

Answer 1

Answer:

(a) $F=968N$

(b) $Ratio=1.542$

Explanation:

Given data

Diameter d=16m

Radius r=d/2=16/2=8.0m

Mass m=64kg

Speed v=11m/s

For Part (a)

From Newtons second law we have:

$F=ma\\as\\a=v^{2}/r\\ so\\F=m\frac{v^{2}}{r}\\ F=64kg\frac{(11m/s)^2}{8m}\\ F=968N$

For Part (b)

The weight is calculated as:

$w=mg\\w=(64kg)(9.81m/s^2)\\w=627.84N$

Therefore we have the ratio of centripetal force to the weight is given as:

$Ratio=F/w\\Ratio=968N/627.84\\Ratio=1.542$