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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

[tex]W_{nc}[/tex]=(1/2[tex]mv_f[/tex]^2)-(1/2[tex]mv_o[/tex]^2) + ([tex]mgh_f[/tex]) – ([tex]mgh_o[/tex])

[tex]W_{nc}[/tex]= ((½ (63kg) [tex](4.4m/s)^2[/tex]) – (½ (63kg) [tex](6.6 m/s)^2[/tex])) + ((63kg) (9.8m/[tex]s^2[/tex])(.802m) + (63kg)(9.8 m/[tex]s^2[/tex]) (0))

[tex]W_{nc}[/tex]= ((½ (63kg) (19.36 [tex]m^2[/tex]/[tex]s^2[/tex])) – ((½ (63kg) (43.56[tex]m^2[/tex]/[tex]s^2[/tex])) + (495.154 J) + 0

[tex]W_{nc}[/tex]= 609.84J - 1372.14J + 495.154J

[tex]W_{nc}[/tex]= -267.146J

Part B

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

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

[tex]W_{nc}[/tex]=(1/2[tex]mv_f[/tex]^2)-(1/2[tex]mv_o[/tex]^2) + ([tex]mgh_f[/tex]) – ([tex]mgh_o[/tex])

[tex]W_{nc}[/tex]= ((½ (63kg) [tex](4.4m/s)^2[/tex]) – (½ (63kg) [tex](6.6 m/s)^2[/tex])) + ((63kg) (9.8m/[tex]s^2[/tex])(.802m) + (63kg)(9.8 m/[tex]s^2[/tex]) (0))

[tex]W_{nc}[/tex]= ((½ (63kg) (19.36 [tex]m^2[/tex]/[tex]s^2[/tex])) – ((½ (63kg) (43.56[tex]m^2[/tex]/[tex]s^2[/tex])) + (495.154 J) + 0

[tex]W_{nc}[/tex]= 609.84J - 1372.14J + 495.154J

[tex]W_{nc}[/tex]= -267.146J

Part B

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