<p>"A child drags his 60N sled at constant speed up a 15 degree hill. He does so by pulling with a 25N force on a rope attached to the sled. If the rope is inclined 35 degrees to the horizontal, (a) what is the coefficient of kinetic friction between sled and snow? (b) At the top of the hill, he jumps on the sled and slides down the hill. What is the magnitude of his acceleration down the slope?"</p>
<p>Correct answers: (a) 0.161 (b) 1.01 m/s^2. I would appreciate help in how to get to these answers.</p>
<p>My current work:</p>
<p>a)
Ffric = uN
since Fnety=0, N=W
thus u=Ffric/W<em>sin(15degrees) [sin15 to get component]
next, since speed is constant there must nto be any a in the x
therefore Ffric=Fpull
rewriting... u = Fpull</em>??? / W*sin(15degrees) = 0.161
My big problem is what is Fpull adjusted by to get this to work out? cos(20) seems logic, but it (and all other cos/sin values of 15 and 35 combined/subtracted/etc don't seem to work).</p>
<p>b) Fnet=ma, thus a=(Fweight-Ffriction)/m=(W * ( sin15 - .161*sin(20)) ) / (W/g)
But that produces 1.99 (not 1.01).</p>
<p>Any help would be greatly appreciated.</p>
<p>ok for part a) the N does not equal Weight cause it's inclined.</p>
<p>BTW I HIGHLY SUGGEST DRAWING A FREE BODY DIAGRAM</p>
<p>the first thing you need to do is componentize the Tension force to level of the hill. since the angle between them is 20 (35-15) The component perpendicular to the hill would be Tsin(20) and the force parallel to the hill would be Tcos(20).</p>
<p>now you need to balance your forces. if you componentize the gravity force (weight). You will get the force perpendicular to the hill as 60cos(15) and force parallel to the hill as 60sin(15).</p>
<p>The normal force is perpendicular to the hill so</p>
<p>let's start balancing the forces perpendicular to the hill (to solve for normal force):
Normal force + Tsin(20) = 60cos(15) /// you know that T is 25
solve for Normal force to get 49.405 N</p>
<p>Normal force = u*Ffric</p>
<p>now you know that the friction force + the parallel component of gravity (60sin(15)) must be equal to the parallel component of the tension force so so</p>
<p>uFfric + 60sin(15) = Tcos(20)
u(49.405) + 60sin(15) = 60cos(20)</p>
<p>u = .161</p>
<p>NOo you've got the wrong work. N does not equal W. N = summation of the foces in the Y, yes. But the forces in the Y are the normal force pushing up and then there is mgcos theta + the y component of the rope's tension acting down. If you plug that in you should be on the right track.</p>
<p>for part b.... is this assuming that the kid has no mass LOL</p>
<p>you've almost got everything right for it</p>
<p>"b) Fnet=ma, thus a=(Fweight-Ffriction)/m=(W * ( sin15 - .161*sin(20)) ) / (W/g)
But that produces 1.99 (not 1.01)."</p>
<p>except that it's not ".161*sin(20)". it should be ".161 * cos(15)". You're trying to find the normal force there, and it should equal the perpendicular component of the gravity</p>