Holding onto a tow rope moving parallel to a frictionless ski slope, a 62.5 kg skier is pulled up the slope, which is at an angle of 8.2° with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at 2.00 m/s and (b) v = 2.00 m/s as v increases at a rate of 0.105 m/s2?(adsbygoogle = window.adsbygoogle || []).push({});

when trying to figure this out, i used the formula,

T = mg sin (theta)

= (62.5)(9.8)(sin 8.2)

= 87.36N

then Fn = mg cos (theta)

=(62.5)(9.8)(cos 8.2)

= 606.2N

then the magnitude came to = 612.5N

for part B....

I used Fn = m(g+a)

= (62.5)(9.8 + .105)

=619.1N

But both of those answers are incorrect... am i going about this problem the wrong way?? any help would be appreciated.

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# Homework Help: Force Problem~

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