During a hockey game on a pond, the defenseman passes a 114-g hockey puck over the ice to the center, who fails to catch it. the puck is traveling at an inital speed of 6.7 m/s. it stops in 18 m due to the frictional force on it from the ice. find the magnitude of the frictional force on the ice

f = 0.1421 N

Explanation:

Kinematics of the hockey puck

The equation of uniformly accelerated rectilinear is:

vf² = v₀²+2ad   Formula (1)

Where:  

v₀: initial  velocity  in m/s  

vf: final  velocity  in m/s  

a: acceleration in m/s²

d: displacement (m)

Data

v₀= 6.7 m/s

vf = 0

d=18 m

Acceleration of the hockey puck

We replace data in the formula (1):

vf² = v₀²+2ad

0 = (6.7)²+2a(18)

- (6.7)² = 2a(18)

- 44.89 = 36a

a = (- 44.89)/36

a = - 1.247 m/s²

Newton's second law for the hockey puck

∑F = m*a

f: frictional force

m= 114 g = 0.114 Kg

-f = (0.114 Kg)* (-1.247 m/s²)

f = 0.1421 N

it is called kinetic energy

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hope this ! : )