This problem has four parts to this. I will list the question for each of the four parts and additional info in the "Description" below.

Question

An ice skater is travelling in a straight, horizontal line on the ice with a velocity of v = 9.5 m/s in the positive x-direction. The coefficient of kinetic friction between the skates and the ice is µ_k = 0.23.

Part a) Write an expression for the normal force, F_N, that the ice pushes against both skates together. Assume the skater's mass is m.

Part b) Using the expression for normal force, write an expression for the skater's acceleration in the x-direction, a.

Part c) How long, t_h in seconds, will it take for the skater's velocity to drop by half?

Part d) How far, d_h in meters, will the skater travel during the time t_h?

Answer 1

Let's go through each part of the problem step by step:

Part a) To find the expression for the normal force (FN) that the ice exerts on the skater, we need to consider the forces acting on the skater in the vertical direction. In this case, the only vertical force is the gravitational force (weight). The normal force FN will be equal in magnitude but opposite in direction to the weight to keep the skater in vertical equilibrium. So:

FN = mg

Where:

FN = Normal force

m = Mass of the skater

g = Acceleration due to gravity (approximately 9.8 m/s²)

Part b) To find the expression for the skater's acceleration in the x-direction (a), we need to consider the horizontal forces acting on the skater. The main horizontal force is the kinetic friction force (Fk) opposing the motion, which is given by:

Fk = µk * FN

Now, we can use Newton's second law to find the acceleration:

ΣF_x = ma

Where:

ΣF_x = Net force in the x-direction (Fk in this case)

m = Mass of the skater

a = Acceleration in the x-direction

So, we have:

µk * FN = ma

Since FN = mg (from Part a), we can substitute it in:

µk * mg = ma

Now, solve for a:

a = (µk * g)

Part c) To find how long it will take for the skater's velocity to drop by half, we can use the following equation of motion:

v = u + at

Where:

v = final velocity (half of the initial velocity, 9.5 m/s / 2 = 4.75 m/s)

u = initial velocity (9.5 m/s)

a = acceleration (from Part b)

t = time (th in this case)

Rearrange the equation:

t = (v - u) / a

Substitute the values:

t = (4.75 m/s - 9.5 m/s) / (µk * g)

Part d) To find how far the skater will travel during the time th, we can use the following equation of motion:

s = ut + (1/2)at²

Where:

s = distance traveled (dh in this case)

u = initial velocity (9.5 m/s)

a = acceleration (from Part b)

t = time (from Part c)

Substitute the values to find dh:

dh = (9.5 m/s) * [(4.75 m/s - 9.5 m/s) / (µk * g)] + (1/2) * (µk * g) * [(4.75 m/s - 9.5 m/s) / (µk * g)]²

Simplify as needed to find the distance.