A40 kg skier skis directly down a frictionless slope angled at l-0o to the horizontal. assume the skier moves in the negative direction of an x axis along the slope. a wind force with component f, acts on the skier. what is f" if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of 1.0 m/s2, and (c) increasing at arate of 2.0 mlsz?

(a) constant. Fx =68,14 N

(b) increasing at a rate of 1.0 m/s2. Fx = 28,14 N

(c) increasing at a rate of 2.0 m/s2. Fx = -11,86 N

Explanation:

In order to solve this situation, it is useful to depict it graphically. in the image we can see drawn the forces that act upon the skier. Once we understand the situation we can proceed and set the equations that model the system. according to Newton’s second law of motion along the negative x axis displayed on the image:

F_x- m*g*cos⁡(80) = -m*a_x → F_x = m*g*cos⁡(80) - m*a_x

Therefore, if we have the scenario where the velocity of the skier is a constant, that means that a_x=0. Replacing on the previous equation, we have:

F_x = 40 kg*9,81 m/s^2 *cos⁡(80) - 40 kg*0 = 68,14 N

In the scenario where the skier’s velocity increases at a rate of 1,0 m/s2, then a_x=1,0 m/s^2 . Replacing on the previous equation, we have:

F_x = 40 kg*9,81 m/s^2 *cos⁡(80) - 40 kg*1,0 m/s^2 = 28,14 N

Finally, in the scenario where the skier’s velocity increases at a rate of 2,0 m/s2, then a_x=2,0 m/s^2 . Replacing on the previous equation, we have:

F_x = 40 kg*9,81 m/s^2 *cos⁡(80) - 40 kg*2,0 m/s^2 = -11,86 N

What this means is that the wind force Fx acts on the opposite direction that the one assumed on the diagram.

Hope everything was clear. If I can help with anything else, let me know. Good day :D

a.   mass number from atomic number

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west