Question

A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.

(a) If its initial speed is 5.8 m/s, what is the magnitude of the frictional force?
(b) What is the coefficient of friction between the puck and the ice?

Answer 1

(a) Ff = 0.128 N

(b μk = 0.1135

Step-by-step explanation:

kinematic analysis

Because the hockey puck moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s

Calculation of the acceleration of the hockey puck

We apply the Formula (1)

vf=v₀+a*t v₀=5.8 m/s , vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t² , d=15.1 m , v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the frictional force (Ff)

We apply Newton's second law

∑F = m*a Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the hockey puck in the attached graphic

∑Fx = m*a m= 115g * 10⁻³ Kg/g = 0.115g , a= -1.12 m/s²

-Ff = 0.115*(-1.115) We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N = 1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135