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

a. 1,188 N

b. 1.836

Step-by-step 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 = (d)/(2)`

`= (16)/(2)`

= 8m

Now the horizontal force is

`F = (m* 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 = (66* 12^2)/(8)`

= 1,188 N

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

`= (1,188 N)/(66\ kg * 9.8m/s^(2))`

= 1.836