Question

A person with a mass of 75 kg skis from rest down a slope at an angle of 31 degrees. The coefficient of kinetic friction of his skis on the snow is 0.095. What is his acceleration?

Answer 1

To calculate the acceleration, we need to determine the net force acting on the skier. This can be done by calculating the parallel force, which is the component of the force of gravity that acts parallel to the slope. Then, subtracting the friction force from the force of gravity gives us the net force. Finally, we can use the equation F = ma to calculate the acceleration.

Step-by-step explanation:

The first step is to calculate the force of gravity acting on the skier. This can be done using the formula:

Force of gravity (Fg) = mass (m) x acceleration due to gravity (g)

Plugging in the values, we get:

Fg = 75 kg x 9.8 m/s² = 735 N

The next step is to determine the component of the force of gravity that acts parallel to the slope. This can be done by multiplying the force of gravity by the sine of the angle of the slope:

Parallel force (Fp) = Force of gravity (Fg) x sin(angle)

Plugging in the values, we get:

Fp = 735 N x sin(31°) = 384.48 N

Finally, we can calculate the net force acting on the skier using the equation:

Net force (Fnet) = Force of gravity (Fg) - Friction force (Ff)

Plugging in the values, we get:

Fnet = 735 N - [(coefficient of kinetic friction) x (normal force)]

Since the skier is moving down the slope and friction opposes the motion, we can set up the equation as:

735 N - [(0.095) x (384.48 N)] = 709.51 N

Therefore, the net force acting on the skier is 709.51 N. Since force and acceleration are directly proportional (Fnet = mass x acceleration), the acceleration of the skier can be calculated using the equation:

Acceleration (a) = Net force (Fnet) / mass (m)

Plugging in the values, we get:

a = 709.51 N / 75 kg = 9.46 m/s²

Therefore, the acceleration of the skier is 9.46 m/s².

Answer 2

4.2 m/s²

Step-by-step explanation:

Draw a free body diagram. There are three forces on the skier:

Weight force mg pulling down

Normal force N perpendicular to the incline

Friction force Nμ parallel up the incline

Sum of the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

mg sin θ − Nμ = ma

Substitute:

mg sin θ − mgμ cos θ = ma

g sin θ − gμ cos θ = a

a = g (sin θ − μ cos θ)

Plug in values:

a = (9.8 m/s²) (sin 31° − 0.095 cos 31°)

a = 4.2 m/s²