After sledding down a hill, my daughter usually wants me to pull her back up the
ID: 2089745 • Letter: A
Question
After sledding down a hill, my daughter usually wants me to pull her back up the hill in the sled as well. The combined weight of Katie and the sled is 290 N, the hill has a 20 degree slop, and the coefficient of kinetic friction between the sled and the snow is Uk=0.40. Assume that I pull her up with a rope that is parallel to the slop. 1.) What is the component of the weight force along the slope? 2.) What is the magnitude of the normal force? 3.) What tension do I have to provide in the rope for Katie to move up hill at a constant speed? If I pull on the rope with a force of 300N, which way will Katie accelerate, and what will her acceleration be?Explanation / Answer
Fg=290, so Fgx=cos(20)(290) which is in the direction of the hill and the normal force will be the component of gravity that is perpendicular to the hill so Fgy=sin(20)(290)=N Now for an object with constant V we know that a=0 so the forces are equal Fp=Fgx+Ff Fgx=272.5 and Ff=? N=.4(99.2) so Fp=692.6N
she will accelerat down the hill F=ma so 692.6-300=29.56a ==> a=21.2968