Solving a Force Problem on Ski Incline | 65

[avb203796]

A person with a mass of 55.0 kg skis down an incline of 40 degrees from the horizontal. If the kinetic coefficient between her skis and the slope is 0.14, what will be her accelaration down the slope? What will be her speed after skiing 8.0 m down the slope if she was at rest when she began her descent? If after reaching the bottom she wishes to climb a slope of 20 degrees at constant velocity, what force must she generate to propel herself up the slope? Assume friction does not change. Also, draw correct vector diagrams that clearly show all forces.

I am unsure how to solve this problem because I do not know how the incline effects forces. I know that on a flat surface there would be a weight force going down and a normal force going straight up and that both would be equal to mg. i also know that there would be a frictional force of kinetic coefficient* mg that would be going horizontally from the back of the skier however, I do not know how all this changes when you are on an incline.

 

[PhY_InTelLecT]

ok..I shall only help you with the first part coz the second part is all bout doing backwards..

Now,.. U have the angle of incline and the force of the object, ie, the person's weight. You can actually make everything look easier by drawing a free body diagram on the person..Draw out the only force and both its horizontal and vertical components. What you need to find now actually is the resultant force that keeps the person accelerating down the slope.. After drawing the free-body diagram, use simple trigonometry to solve for your horizontal and vertical force components ./ The vertical force component will be the same as the normal force that u will nid in calculating friction. The horizontal force component will be the forward force cause by the weight of the person..

Finding the difference between this 2 forces will give you the resultant force on the person.. Then by then i think it shld be quite easy alr. Juz find acceleration from the resultant force//.

 

[andrevdh]

The skier experiences three forces, her weight downwards, the normal force of the slope, which is perpendicular to the incline and the frictional force which is parallel to the incline pointing up the slope (friction always works along the surface). The confusing one is the normal force - it is the supporting force of the surface, the snow in this case. The total force of the surface is therefore the sum of the normal and the frictional force, which will therefore point at some odd angle leaning a bit upwards from the perpendicular to the slope. If there were no friction then the total force coming from the snow will be just perpendicular to the slope.

 

[PhY_InTelLecT]

The skier only experiences two forces..Friction and the normal force by the surface.. Weight is a force exerted by the skier on the surface.. there's a different..
This 2 force the skier experiences are indirect 'reactions' of the the components of the force, weight.