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 information on the distance.
You cant solve this without knowing height of the hill
To find the time it takes for the skier to go from rest at the top of the inclined plane to rest on the flat ground, we can use the principles of physics and apply them to this situation.
First, let's analyze the forces acting on the skier as they move down the inclined plane. There are two main forces involved:
1. Gravitational force (Fg): This force acts vertically downwards and can be calculated using the formula Fg = m * g, where m is the mass of the skier and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. Friction force (Ffriction): This force acts parallel to the inclined plane and opposes the motion of the skier. According to the problem, the coefficient of friction is 0.100.
Now, let's break down the problem into two parts: the skier moving down the inclined plane and the skier moving on the flat ground.
1. Skier moving down the inclined plane:
On the inclined plane, the net force acting on the skier is the component of the gravitational force that is parallel to the plane minus the friction force:
Fnet = Fg * sin(theta) - Ffriction
= m * g * sin(theta) - (coefficient of friction) * m * g * cos(theta)
The acceleration (a) of the skier on the inclined plane can be determined using Newton's second law:
Fnet = m * a
So, by substituting the values, the equation becomes:
m * a = m * g * sin(theta) - (coefficient of friction) * m * g * cos(theta)
Simplifying, we get:
a = g * (sin(theta) - (coefficient of friction) * cos(theta))
2. Skier moving on the flat ground:
Once the skier reaches the flat ground, the inclined plane is no longer exerting a force on the skier. Therefore, the net force acting on the skier is only due to friction, opposing the skier's motion:
Fnet = - Ffriction
Using Newton's second law again, we have:
m * a = - (coefficient of friction) * m * g
Simplifying, we get:
a = - (coefficient of friction) * g
As the skier comes to rest on the flat ground, its final velocity (vf) is 0. We can use the kinematic equation:
vf = vi + a * t
Since the skier starts from rest, the initial velocity (vi) is 0.
Therefore, the equation becomes:
0 = 0 + (- (coefficient of friction) * g) * t
Simplifying, we have:
- (coefficient of friction) * g * t = 0
Now, we can solve for the time (t):
t = 0 / (-(coefficient of friction) * g)
Since the denominator is a negative value, we can ignore the negative sign.
Finally, substituting the value of g (approximately 9.8 m/s^2) and the given coefficient of friction (0.100), we can calculate the time it takes for the skier to go from rest on the inclined plane to rest on the flat ground:
t = 0 / (-(0.100) * 9.8)
t = 0 / (-0.980)
t = 0
Therefore, the skier takes no time to go from rest on the inclined plane to rest on the flat ground.