A 75 kg skier starts at rest on top of a icy inclined plane (with a 30 degree angle to the ground). He ends up going past the inclined plane and onto a flat ground. And comes to a rest. The coefficient of friction is .100. How long does it take for the skier to go from rest (at the top of the inclined plane) to rest (on the flat ground)?
There is no more information available. I need help solving it.
We need to know the length of the hill.
Length = 100 m?
A = 30o
m = 75 kg
h = 100*sin30 = 50 m. = Ht. of hill.
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*50 = 980
V = 31.3 m/s At bottom of hill.
V^2 = Vo^2 + 2a*d
a=(V^2-Vo^2)/2d=(31.3^2-0)/200=4.9 m/s^2
V = Vo + a*t = 31.3
0 + 4.9*t = 31.3
T1 = 6.39 s. = Time to reach gnd.
V = Vo + a*t = 0
31.3 - 4.9t = 0
4.9t = 31.3
T2 = 6.39 s. on level gnd.
T = T1 + T2 = 6.39 + 6.39 = 12.78 s.
To solve this problem, we can break it down into two parts: the skier's motion on the inclined plane and the skier's motion on the flat ground.
On the inclined plane, the skier experiences two forces: the gravitational force pulling him down the incline and the frictional force opposing his motion. The gravitational force is given by the equation Fgrav = m * g * sin(theta), where m is the mass of the skier, g is the acceleration due to gravity (approximately 9.8 m/s^2), and theta is the angle of the incline (30 degrees in this case). The frictional force is given by the equation Ffrict = u * m * g * cos(theta), where u is the coefficient of friction.
Now, let's find the acceleration of the skier on the inclined plane using Newton's second law, Fnet = m * a, where Fnet is the net force acting on the skier. On the inclined plane, the net force is the difference between the gravitational force and the frictional force, so we have Fnet = Fgrav - Ffrict. And the acceleration a is given by a = Fnet / m.
Next, we need to determine the time it takes for the skier to reach the bottom of the inclined plane. We can use the kinematic equation, vf = vi + a * t, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. Since the skier starts from rest (vi = 0), the equation simplifies to vf = a * t.
Once the skier reaches the flat ground, he will experience a negative acceleration due to the friction force acting against his motion. We can use the equation vf^2 = vi^2 + 2 * a * d, where vf is the final velocity (0 m/s), vi is the initial velocity (the velocity at the bottom of the incline), a is the acceleration (negative due to the friction force), and d is the distance traveled on the flat ground. Solving for d, we have d = -vi^2 / (2 * a).
Now, we can use the equations we derived to solve for the time it takes for the skier to go from rest on the inclined plane to rest on the flat ground.
1. Calculate the gravitational force: Fgrav = 75 kg * 9.8 m/s^2 * sin(30 degrees) = 367.5 N
2. Calculate the frictional force: Ffrict = 0.100 * 75 kg * 9.8 m/s^2 * cos(30 degrees) = 66.9 N
3. Calculate the net force: Fnet = Fgrav - Ffrict = 367.5 N - 66.9 N = 300.6 N
4. Calculate the acceleration on the inclined plane: a = Fnet / m = 300.6 N / 75 kg = 4 m/s^2
5. Calculate the time to reach the bottom of the inclined plane: vf = a * t -> 0 m/s = 4 m/s^2 * t -> t = 0 s
6. Calculate the initial velocity at the bottom of the incline using the kinematic equation: vf = vi + a * t -> 0 m/s = vi + 4 m/s^2 * 0 s -> vi = 0 m/s
7. Calculate the distance traveled on the flat ground: d = -vi^2 / (2 * a) -> d = -0 m/s / (2 * -4 m/s^2) -> d = 0 m
Therefore, the skier goes from rest on the inclined plane to rest on the flat ground instantaneously (taking no time).