Question

A skier starts from rest at the top of a hill that is inclined at 9.5° with the horizontal. The hillside is 155 m long, and the coefficient of friction between snowand skis is 0.0750. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontalportion of the snow before coming to rest?
____ m

Answer 1

The acceleration of the skier down the incline is calculated using Newton's 2nd Law of Motion, i.e., F = Ma
where
F = frictional force
M = mass of the skier
a = acceleration
By definition,
F = (mu)(N)
where
mu = coeff of friction = 0.075 (given)
N = normal force = Mg(cos 9.5)
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
Therefore,
0.075(Mg)(cos 9.5) = Ma
Since "M" appears on both sides of the equation, it will simply cancel out hence the above is simplified to
0.075(9.8)(cos 9.5) = a
a = 0.7249 m/sec^2
Next formula to use is
Vf^2 - Vo^2 = 2as
where
Vf = velocity of the skier at the bottom of the hill
Vo = initial velocity of skier = 0 (since skier started from rest)
a = acceleration = 0.7249 m/sec^2 (as calculated above)
s = distance travelled by skier along the incline = 155 m (given)
Substituting values,
Vf^2 - 0 = 2(0.7249)(155)
=>Vf = 15 m/sec
At the bottom of the hill and going to the portion where the snow is level, the acceleration of the skier is computed as (still using Newton's 2nd Law ofMotion),
F = Ma and since
F = (mu)(N) = (mu)(Mg)
and therefore,
(mu)(Mg) = Ma
and again, since M appears on both sides of the equation, it will simply cancel out hence, the above becomes
a = (mu)(g) 0.075*9.8
a = 0.735 m/sec^2
The last formula to use is
Vf^2 - Vo^2 = 2as
where
Vf = final velocity of skier = 0 (when skier comes to rest)
Vo = velocity of skier at bottom of hill = 15 m/sec (as calculated above)
a = 0.735 m/sec^2
s = gliding distance of skier along horizontal portion
Substituting values,
0 - 15^2 = 2(-0.735)(s)
NOTE the negative sign attached to the acceleration. This simply denotes that the skier is slowing down during the glide along the horizontalportion.
Solving for "s",
=>s = 15.^2/(2 * 0.735)
=>s = 153.06 m

Answer 2

net, height of the hill = 155 sin 9.5 = 25.58 m
now, let mass of skier = m
so, till the slope of the hill, frictional force acting F= 0.0750 x N(normal force)
=0.075 x M x gcos9.5 = 0.7397M
now, work done by friction till the slope of hill = F x distance = 0.7397M x 155= 114.656M----------------------------------------------------------------------------------1.

now, let that it goes to a distance k, gliding, after the slope of hill.
So, frictional work till that distance k = 0.075 x N(=Mg) x k = 0.75M x k.------------2.
now, term 1. + term 2. = total potential energy at top of hill.
So, 114.656M + 0.75Mk = Mg x 25.58(height of hill)
or, 114.656 +0.75k=255.8
or, 0.75k=141.144
so, k= distance of horizontal portion before coming to rest= 188.2 m

Answer 3

using the enrgy,

PE at top oh hill = mgh = m(9.8) 155sin9.5( 155 sin9.5 is the height of hill)

energy lost due to friction =μmgcosθ( 155)

KE at bottom of hilll = PE - enrgy lost due tyo friction

0.5mv^2 = m(9.8)155sin 9.5 - 0.0750( m)(9.8)cos9.5(155)

solving for v^2

0.5v^2= 250.707 - 112.362

v^2 = 276.69 apprx

a(retardation ) on ground =μg = 0.0750 ( 9.8) =- 0.735 m/s^2

using kineamatic equation,

V^2 - v^2 = 2as

0 - 276.69 = 2(- 0.735)s

s(displacemnt) = 188.22 m apprx