Answer: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
