A skater with an initial speed of 5.90 m/s stops propelling himself and begins to coast across the ice, eventually coming to rest. Air resistance is negligible. (a) The coefficient of kinetic friction between the ice and the skate blades is 0.0500. Find the deceleration caused by kinetic friction. (b) How far will the skater travel before coming to rest

a) - 0.4905 m/s²   b) distance = 35.48 m

Explanation:

Given that  

The initial velocity of the skater = 5.90 m/s

 kinetic friction coefficient = 0.0500

final velocity = 0 m/s(since it  comes to rest)

deceleration cause by the kinetic friction = ?

we know that  

F = μN

and N= mg

Therefore;

F = μ m g(1)

also  that

F = m a(2)

with our common Force, F, equating  (1) and (2), we have that

m a = - μ m g

a = - μ g

a = - 0.05 × 9.81

a = - 0.4905 m/s²

The deceleration cause by the kinetic friction is a = - 0.4905 m/s²

b)

The distance the skater  travels before stopping

is given as

    Vf² = v₀² - 2 a x

  final velocity = 0 m/s(since it  comes to rest)  

Therefore We have that

 0 = v₀² - 2 a x

 x = - v₀² / 2 a

 x = 5.90² / (2 x 0.4905  )

34.81/0.981  

x = 35.48 m

Or

using

v²-u² = 2aS final velocity = 0 m/s(since it  comes to rest)  

0²-5.90² = -2×0.4905×S

34.81=0.981S

S= 34.81/0.981

S=35.48m

he kicked it 10 feet straight up into the air? i think this works.

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