# Work and Energy of skier

• chmilne

#### chmilne

I believe I have part A correct, but I have having some trouble with part B.

A 63-kg skier coasts up a snow-covered hill that makes an angle of 25° with the horizontal. The initial speed of the skier is 6.6 m/s. After coasting a distance of 1.9 m up the slope, the speed of the skier is 4.4 m/s. (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

Part A

Part A

$$W_{nc}$$=(1/2$$mv_f$$^2)-(1/2$$mv_o$$^2) + ($$mgh_f$$) – ($$mgh_o$$)
$$W_{nc}$$= ((½ (63kg) $$(4.4m/s)^2$$) – (½ (63kg) $$(6.6 m/s)^2$$)) + ((63kg) (9.8m/$$s^2$$)(.802m) + (63kg)(9.8 m/$$s^2$$) (0))
$$W_{nc}$$= ((½ (63kg) (19.36 $$m^2$$/$$s^2$$)) – ((½ (63kg) (43.56$$m^2$$/$$s^2$$)) + (495.154 J) + 0
$$W_{nc}$$= 609.84J - 1372.14J + 495.154J
$$W_{nc}$$= -267.146J

Part B

Because the skier is going up hill, I would think that the angle for the formula W = (Fcos$$\theta$$)s is 15$$5^o$$. So,

-267.146J = (Fcos(155))1.9m
-140.60316 = Fcos(155)
-140.60316 = -.9063078F
155.136N = F

This cannot be true because the professor has mentioned that, to 1 significant digit, the answer is 100N. I'm to find the anser to 2 SF, so, 155.136 would not be true.

Thanks,
Tony

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The force of friction acts along the full displacemnt of 1.9 m, not a componant of this.

I get it. The angle should be 180, not 155 because the force run along the x-axis and using 155 would not be the entire x axis. This mean that the answer to part b is 140.603N. Correct?

I worked this problem and obtained 155N. So I assumed that the problem
means a 1.9m "vertical" rise and obtained the desired answer of 100N.

The problem states the skier is on a hill that is 25 degrees from the horizontal and that the displacement is 1.9m. So, the 1.9m could not be veritcle, it would also be 25 degrees from the horizontal. However, to answer part b, you need to calculate for 180 degrees, not 155 degrees.

chmilne said:
I get it. The angle should be 180, not 155 because the force run along the x-axis and using 155 would not be the entire x axis. This mean that the answer to part b is 140.603N. Correct?
Correct (but please, drop those insig digits! It's fingernails on the chalkboard to me!), I get 140 N .

Easier way to think of it; the force vector and the displacement vector are parallel , therefore no vector resolution is required.

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A simple work and energy Q

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I tried to draw the q but could not succed Chi Meson said:
Easier way to think of it; the force vector and the displacement vector are parallel , therefore no vector resolution is required.
Another way to think of it is to remember what a dot product is. chmilne, you start off with the formula $$W = F\cos{\theta}s$$, but a more general form is $$W = \vec{F}\cdot \Delta \vec{s}$$.

If you think of it in terms of dot products and vector resolution, then you can mentally and graphically verify that you're selecting the correct angle.

(An even more general form would be $$W = \int_{x_1}^{x_2} \vec{F} \,d\vec{x}$$, but that's hardly needed at elementary mechanics-level physics.)