Question 7 (5 points)
blaine steps onto a ski slope with an angle of 25°. there is a coefficient of kinetic friction of
0.75 between him and the ground. he has a mass of 65 kg.
b. write the expressions for net force in the x and y directions be sure to tilt your axis along the incline
c. calculate the net force and acceleration in the x and y directions

The net force is 165N an the the acceleration is 2.5 m/s2

Explanation: First find the component of force with respect to the angle of inclination.

F = mgsin (theta) = 65 kg x 9.8 m/s2 x sin (25°) = 286N

Next find the value of the normal force

N = mgcos ( theta)

= 65 kg x 9.8m/s² x cos (25°)

= 577N

Substitute the Value of N to find the frictional force.

fk= Coeffiecient of kinetic friction x N

= 0.75x 577 N = 433 N

Solve for the net force through summation of forces.

433 N - 268N = 165N

To find a. Use F= ma

Derive so that a= F/m

a= 165N / 65kg

= 2.54 m/s2

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