Question

A skier is moving down a snowy hill with an acceleration of 0.40 m/s^2. The angle of the slope is 5.0° to the horizontal. What is the acceleration of the same skier when she is moving down a hill with a slope of 19°? Assume the coefficient of kinetic friction is the same in both cases.

Answer 1

The acceleration of the skier when moving down a hill with a slope of 19° will be the same as the acceleration when moving down a slope of 5.0°, which is 0.40
`m/s^2.`

Step-by-step explanation:

To find the acceleration of the skier moving down a hill with a slope of 19°, we can use the formula:

acceleration = g * sin(theta) - friction

Since the coefficient of kinetic friction is the same and we can neglect air resistance, the friction term will be the same in both cases. The only difference will be the angle theta.

So, the acceleration of the skier when moving down a hill with a slope of 19° will be the same as the acceleration when moving down a slope of 5.0°, which is 0.40 m/s^2.

Answer 2

ax=2.76 m/s²

Step-by-step explanation:

Conceptual analysis

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

a=0.40 m/s²

β =5°

g = 9.81 m/s² : acceleration due to gravity

W= m*g

x-y weight components

Wx= Wsinβ= m*g*sin5°

Wy= Wcosβ=m*g*cos5°

Friction force : Ff

Ff= μk*N

μk: coefficient of kinetic friction

N : Normal force (N)

Problem development

look at the skier's free body diagram in the attached graphic

We apply the formula (1)

∑Fy = m*ay , ay=0

N-Wy = 0

N = Wy

N= m*g*cos5°

∑Fx = m*ax

Wx-Ff= m*ax

m*g*sin5° -μk*m*g*cos5° = m*0.4 : we divide by m on both sides of the equation:

g*sin5°- μk*g*cos5°= 0.4

g*sin5° - 0.4 = μk*g*cos5°

μk = (9.8*sin5° - 0.4 ) /( 9.8*cos5°)

μk = 0.0465

Acceleration of the same skier when she is moving down a hill with a slope of 19°, μk = 0.0465

∑Fx = m*ax

Wx-Ff= m*ax Wx=m*g*sin 19°, N=*m*g*cos19°, Ff=μk*N

m*g*sin 19°-(0.0465)*(m*g*cos19°) = m*ax we divide by m , and g= 9.8

9.8*sin 19°-(0.0465)*(9.8*cos19°) = ax

ax=2.76 m/s²