Answer :
The acceleration of the ski up the 75-degree inclined track, given a 141 N applied force parallel to the surface and a 31.5 N frictional force, is approximately [tex]\(-9.59\) m/s\(^2\)[/tex], directed up the incline. This calculation uses Newton's second law of motion considering the net force acting along the incline to determine the resulting acceleration.
To solve for the acceleration of the ski up the inclined track under the given conditions, we'll use Newton's second law of motion along the direction of the incline. The forces acting on the ski parallel to the track surface include the applied force F = 141 N, the frictional force f = 31.5 N, and the component of gravitational force acting down the incline [tex]\( mg \sin(\theta) \)[/tex], where [tex]\( \theta = 75^\circ \)[/tex] is the angle of inclination.
First, resolve the gravitational force into two components:
[tex]\[ F_{\text{gravity, parallel}} = mg \sin(\theta) \][/tex]
where [tex]\( g = 9.8 \) m/s\(^2\)[/tex] is the acceleration due to gravity and m = 220 kg is the mass of the ski.
The net force parallel to the incline [tex]\( F_{\text{net, parallel}} \)[/tex] is:
[tex]\[ F_{\text{net, parallel}} = F \cos(\theta) - f - F_{\text{gravity, parallel}} \][/tex]
Substitute the values:
[tex]\[ F_{\text{net, parallel}} = 141 \cos(75^\circ) - 31.5 - (220 \times 9.8 \times \sin(75^\circ)) \][/tex]
Now, use Newton's second law [tex]\( F_{\text{net, parallel}} = ma \)[/tex] to solve for acceleration ( a ):
[tex]\[ ma = 141 \cos(75^\circ) - 31.5 - (220 \times 9.8 \times \sin(75^\circ)) \][/tex]
Solve for ( a ):
[tex]\[ a = \frac{141 \cos(75^\circ) - 31.5 - (220 \times 9.8 \times \sin(75^\circ))}{220} \][/tex]
After calculating the values:
[tex]\[ a \approx \frac{141 \times 0.2588 - 31.5 - (220 \times 9.8 \times 0.9659)}{220} \][/tex]
[tex]\[ a \approx \frac{36.4918 - 31.5 - 2114.808}{220} \][/tex]
[tex]\[ a \approx \frac{-2109.8162}{220} \][/tex]
[tex]\[ a \approx -9.5901 \text{ m/s}^2 \][/tex]
Therefore, the acceleration of the ski up the inclined track is approximately [tex]\( \boxed{-9.59} \) m/s\(^2\)[/tex], directed up the incline. The negative sign indicates that the acceleration is in the direction opposite to the applied force.
