A skier of mass 60 kg skies down a 35° slope; the coefficient of friction between her skis and the snow is 0.40
(a) Calculate the skier’s acceleration.
(b) If the skier started at 5 m/s, how fast is she going after travelling 25 m down the slope?
mu = 0.40, good grief wax those skis !!!!
normal force on snow = m g cos 35
so friction force up slope = mu m g cos 35=0.4*60*9.81*0.819 = 193 N
weight component down slope = m g sin 35 = 338 N
F = m a
338 - 193 = 60 a
a = 2.41 m/s^2
v = Vi + a t
v = 5 + 2.41 t
d = Vi t +(1/2) a t^2
d = 5 t + 1.2 t^2 = 25
1.2 t^2 + 5 t - 25 = 0
t =2.93 seconds
then v = 5 + 2.41 * 2.93 = 12.1 m/s
Bless you anonymous, thank you so much!
To solve this problem, we can use Newton's second law of motion. The force acting on the skier can be broken down into two components: the force of gravity down the slope and the force of friction.
(a) To calculate the skier's acceleration:
1. Resolve the force of gravity into two components: one parallel to the slope (mg sinθ) and one perpendicular to the slope (mg cosθ), where θ is the angle of the slope (35°).
F_parallel = mg sinθ = (60 kg) * (9.8 m/s^2) * sin35°
F_parallel = (60 kg) * (9.8 m/s^2) * 0.573576436351046
2. Calculate the force of friction using the coefficient of friction (μ) and the perpendicular force of gravity.
F_friction = μ * F_perpendicular = (0.40) * [mg cosθ = (60 kg) * (9.8 m/s^2) * cos35°]
F_friction = (0.40) * (60 kg) * (9.8 m/s^2) * 0.8191520442889918
3. Calculate the net force acting on the skier.
F_net = F_parallel - F_friction
4. Use Newton's second law of motion (F = ma) to calculate acceleration.
F_net = ma
a = F_net / m
(b) To calculate the skier's final velocity after traveling 25 m down the slope:
1. Use the equation of motion:
vf^2 = vi^2 + 2ad
where
vf = final velocity
vi = initial velocity (5 m/s)
a = acceleration (calculated in part a)
d = displacement (25 m)
2. Substitute the known values into the equation and solve for vf.
To answer these questions, we can use the principles of Newtonian mechanics. Specifically, we can use the equations of motion to analyze the skier's motion down the slope.
(a) To calculate the skier's acceleration, we need to find the net force acting on the skier. The net force can be expressed as the sum of the gravitational force and the force of friction.
The gravitational force acting on the skier is given by: F_gravity = m * g * sin(θ),
where m is the mass of the skier, g is the acceleration due to gravity (approximately 9.8 m/s^2), and θ is the angle of the slope (35°).
The force of friction can be calculated using: F_friction = μ * F_normal,
where μ is the coefficient of friction and F_normal is the normal force. The normal force is the component of the gravitational force perpendicular to the slope and can be expressed as: F_normal = m * g * cos(θ).
Substituting these values into the equation for net force:
F_net = F_gravity - F_friction = m * g * sin(θ) - μ * m * g * cos(θ).
Finally, the acceleration can be calculated using Newton's second law: F_net = m * a.
So, a = (m * g * sin(θ) - μ * m * g * cos(θ)) / m,
which simplifies to: a = g * (sin(θ) - μ * cos(θ)).
Now let's substitute the given values:
m = 60 kg,
g = 9.8 m/s^2,
θ = 35°,
μ = 0.40.
Plugging these values into the equation, we get:
a = 9.8 * (sin(35°) - 0.40 * cos(35°)).
Solving this equation will give us the value of acceleration.
(b) To determine the skier's final velocity after traveling 25 m down the slope, we can use the equation of motion:
v^2 = u^2 + 2as,
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Given the initial velocity (u = 5 m/s) and the acceleration (calculated in part (a)), we can calculate the final velocity (v) by substituting these values into the equation and solving for v:
v^2 = (5 m/s)^2 + 2 * a * 25 m.
Solving this equation will give us the square of the final velocity. To find the actual final velocity, we take the square root of the result.