a 68kg skier is speeding down a ski slope that makes a 32 angle with the horizontal. The co-efficient of friction between the skies and the snow is 0.05. What is the acceleration of the skier?
I want to understand how to do this
m*g = 68kg * 9.8N./kg = 666.4 N. = Wt. of skier.
Fp = 666.4*sin32 = 353.1 N. = Force parallel with incline.
Fn = 666.4*cos32 = 565.1 N. = Normal =
Force perpendicular to the incline.
Fk = u*Fn = 0.05 * 565.1 = 28.26 N. =
Force of kinetic friction.
Fp-Fk = m*a
a=(Fp-Fk)/m =(353.1-28.26)/68=4.78m/s^2
Note: Fp-Fk = Net force.
To solve this problem, we need to take into account the forces acting on the skier. There are two significant forces: the force of gravity pulling the skier downward and the force of friction opposing the skier's motion.
The force of gravity 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/s²).
Fg = (68 kg) * (9.8 m/s²) = 666.4 N
The force of gravity can be broken down into two components: one parallel to the slope (Fgsinθ) and one perpendicular to the slope (Fgcosθ). Theta (θ) is the angle between the slope and the horizontal.
The component parallel to the slope (Fgsinθ) is responsible for accelerating the skier downhill, and it is opposed by the force of friction.
The force of friction can be calculated by multiplying the coefficient of friction (μ) by the normal force (Fn). The normal force is the force exerted by a surface perpendicular to it.
Fn = Fgcosθ = (666.4 N) * cos(32°) ≈ 562.4 N
The force of friction (Ff) is then given by Ff = μ * Fn.
Ff = (0.05) * (562.4 N) = 28.1 N
The net force acting on the skier in the downhill direction is the difference between the force of gravity parallel to the slope (Fgsinθ) and the force of friction (Ff).
Net force = Fgsinθ - Ff
The acceleration (a) of the skier can be obtained using Newton's second law: F = ma, where F is the net force acting on the skier.
a = (Fgsinθ - Ff) / m
a = ((666.4 N) * sin(32°) - 28.1 N) / 68 kg
a ≈ 3.62 m/s²
Therefore, the acceleration of the skier on the slope is approximately 3.62 m/s².