A 55 kg ice skater is at rest on a flat skating rink. A 198 N horizontal forceis needed to set the skater in motion.However, after the skater is inmotion, a horizontal force of 175 N keeps the skater moving at a constantvelocity. Find the coefficients of static and kinetic friction between theskates and the ice.
Static mu = 198N/(55 kg*9.8m/s^2)
Kinetic mu = 175/(55*9.8)
Must be pretty rough ice. Get out the Zambomi.
static = 0.37
kinetic = 0.32
Given :
M= 55kg Fsmax= 198N.
Kinetic friction= 175
Unknown:
Coefficients of kinetic friction = ?
Coefficient of static friction = ?
Coefficient of kinetic friction:
Kinetic friction (Fk) / Normal friction (Fn)
(Fn=mg)
Therefore: 175 / (55)(9.81) = 0.32
Coefficient of static friction:
Fsmax / Normal friction (Fn)
Therefore: 198 / (55)(9.81) = 0.37
Good luck on exams🤝
Thank you!
Well, well, well, it seems like this ice skater has quite the forceful dilemma! To find the coefficients of static and kinetic friction, we need to do a little bit of math.
First, let's determine the force required to set the skater in motion. The horizontal force needed is 198 N, so that means the static friction force is equal to 198 N. Now, we can calculate the coefficient of static friction (μs) using the formula:
μs = Fs / N
where Fs is the static friction force, and N is the normal force.
Since our skater is at rest, we can find the normal force by multiplying the mass of the skater (55 kg) by the acceleration due to gravity (9.8 m/s^2):
N = m * g = 55 kg * 9.8 m/s^2 = 539 N
Plugging in the values, we get:
μs = 198 N / 539 N ≈ 0.3668
So, the coefficient of static friction between the skates and the ice is approximately 0.3668.
Now, onto the coefficient of kinetic friction (μk). Once the skater is in motion, a horizontal force of 175 N keeps the skater moving at a constant velocity. This is the force of kinetic friction (Fk). We can use the following formula to find the coefficient of kinetic friction:
μk = Fk / N
Using the same normal force (N) as before, we have:
μk = 175 N / 539 N ≈ 0.3243
Therefore, the coefficient of kinetic friction between the skates and the ice is approximately 0.3243.
So there you have it! The coefficient of static friction is around 0.3668, and the coefficient of kinetic friction is about 0.3243. But remember, even though friction can be a bit of a drag, it still helps this skater glide on the ice!
To find the coefficients of static and kinetic friction between the skates and the ice, we can use the equations related to force and friction.
1. The force required to set the skater in motion is the maximum static friction force, which can be calculated using the formula:
\(f_{\text{static max}} = \mu_{\text{static}} \cdot N\)
where \(f_{\text{static max}}\) is the maximum static friction force, \(\mu_{\text{static}}\) is the coefficient of static friction, and \(N\) is the normal force.
2. The force needed to keep the skater moving at a constant velocity is the kinetic friction force, which can be calculated using the formula:
\(f_{\text{kinetic}} = \mu_{\text{kinetic}} \cdot N\)
where \(f_{\text{kinetic}}\) is the kinetic friction force, \(\mu_{\text{kinetic}}\) is the coefficient of kinetic friction, and \(N\) is the normal force.
Given:
Mass of the skater (\(m\)): 55 kg
Force to set the skater in motion (\(f_{\text{static max}}\)): 198 N
Force to keep the skater moving (\(f_{\text{kinetic}}\)): 175 N
To find the coefficients of static and kinetic friction, we need to find the normal force (\(N\)) acting on the skater.
The normal force is equal to the weight of the skater, which can be calculated using the formula:
\(N = m \cdot g\)
where \(g\) is the acceleration due to gravity (approximately 9.8 m/s²).
Let's calculate the normal force:
\(N = 55 \, \text{kg} \cdot 9.8 \, \text{m/s²} = 539 \, \text{N}\)
Now, we can calculate the coefficients of static and kinetic friction.
Coefficient of static friction (\(\mu_{\text{static}}\)):
Using the equation for maximum static friction force:
\(f_{\text{static max}} = \mu_{\text{static}} \cdot N\)
we can rearrange the equation to solve for \(\mu_{\text{static}}\):
\(\mu_{\text{static}} = \frac{f_{\text{static max}}}{N}\)
Substituting the values:
\(\mu_{\text{static}} = \frac{198 \, \text{N}}{539 \, \text{N}} \approx 0.367\)
Coefficient of kinetic friction (\(\mu_{\text{kinetic}}\)):
Using the equation for kinetic friction force:
\(f_{\text{kinetic}} = \mu_{\text{kinetic}} \cdot N\)
we can rearrange the equation to solve for \(\mu_{\text{kinetic}}\):
\(\mu_{\text{kinetic}} = \frac{f_{\text{kinetic}}}{N}\)
Substituting the values:
\(\mu_{\text{kinetic}} = \frac{175 \, \text{N}}{539 \, \text{N}} \approx 0.325\)
Therefore, the coefficients of static and kinetic friction between the skates and the ice are approximately 0.367 and 0.325, respectively.