Respuesta :
Answer:
stopping distance = 38.75 m
Step-by-step explanation:
Given data:
height of sliding hill = 3.1 m
slope =25 degree
coefficient of friction is 0.08
Gravitational energy is given as E
E = Mgh
E = 3.1 Mg joule
As the hill is frictionless, thus all energy is converted to kinectic energy
Total work done to stop the sliding is
work done = force × distance
we know that
coefficient of friction is given as
[tex]\mu = \frac{F_f}{F_n}[/tex]
where F_f - friction force
F_n - normal friction force
[tex]F_f = 0.08 \times Mg[/tex]
so wrok done [tex] = 0.08 Mg \times stopping/ distance[/tex]
[tex]3.1 Mg = 0.08 Mg \times stopping / distance[/tex]
so
stopping distance = 38.75 m
Answer:
38.8 m
Step-by-step explanation:
Height of hill, h = 3.1 m
slope of hill, θ = 25°
coefficient of friction, μ = 0.08
Let v be the velocity of Josh as he reaches at the bottom of hill.
By use of conservation of energy
m x g x h = 0.5 x m x v²
9.8 x 3.1 = 0.5 x v²
v = 7.8 m/s
Now Josh is moving on the snow with initial velocity, u = 7.8 m/s
final velocity is zero. Let he travels upto a distance of s.
Use third equation of motion
v² = u² + 2as
0 = 7.8² - 2 x 0.08 x 9.8 x s
s = 38.8 m
Thus, the distance traveled by Josh before stopping is 38.8 m.
