Question

An ice skater is traveling in a straight, horizontal line on the ice with a velocity of v= 7.5 m/s in the positive x-direction. The coefficient of kinetic friction between the skates and the ice is uk= 0.22.

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

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

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

d) How far, dh in meters, will the skater travel during the time th?

Please answer with as much explanation as possible, Thank you!

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the ice skater = m

Normal force on the ice skater = F_N

The normal force on the ice skater supports the weight of the ice skater.

F_N = mg

Coefficient of kinetic friction between the skates and the ice = _k = 0.22

Friction force on the skater = f

f = -_kF_N (Negative as it is acting along the negative X-direction)

f = -_kmg

Acceleration of the ice skater = a

ma = f

ma = -_kmg

a = -_kg

a = -(0.22)(9.81)

a = -2.158 m/s²

Initial velocity of the ice skater = V₁ = 7.5 m/s

Final velocity of the ice skater = V₂ = V₁/2 = 3.75 m/s

Time taken to half the ice skater's velocity = T_h

V₂ = V₁ + aT_h

3.75 = 7.5 + (-2.158)T_h

T_h = 1.738 sec

Distance traveled by the skater in time 'T_h' = D_h

D_h = V₁T_h + aT_h²/2

D_h = (7.5)(1.738) + (-2.158)(1.738)²/2

D_h = 9.77 m

a) Normal force = F_N = mg

b) Acceleration of the skater = a = -_kg

c) Time taken for the skater's velocity to drop by half = T_h = 1.738 sec

d) Distance traveled by the skater in time 'T_h' = D_h = 9.77 m