Question

In short-track speed skating, the track has straight sections and semi-circles 16 m in diameter. Assume that a 66 kg skater goes around the turn at a constant 12 m/s.
A. What is the horizontal force on the skater?
B. What is the ratio of this force to the skater's weight?

Answer 1

Answer:

a. 1,188 N

b. 1.836

Explanation:

The computation is shown below:

a. For horizontal force, first we need to find out the circular path radius which is shown below:

As we know that

$r = \frac{d}{2}$

$= \frac{16}{2}$

= 8m

Now the horizontal force is

$F = \frac{m\times v^2}{r}$

where,

m = 66 kg

v = 12 m/s

and r = 8 m

Now placing these values to the above formula

So, the horizontal force is

$F = \frac{66\times 12^2}{8}$

= 1,188 N

b. Now the ratio of force to the weight of skater is

$= \frac{1,188 N}{66\ kg \times 9.8m/s^{2}}$

= 1.836