You are pushing a sled in which your little sister is seated up a 27° slope (one that makes an angle of 27° with the horizontal). If the mass of the sled and girl is 50 kg, and the coefficient of kinetic friction between the sled and the surface is 0.100, with what magnitude of force do you need to push in order for the sled to move with constant velocity?
+37084 First figure out the downslope and 'normal' forces of the 50kg mass
Downslope = 50 sin27 = 22.69
Normal = 50 cos27 = 44.55 kg
Since you are pushing the box UP the hill the force of friction is acting against you ...down the hill
coeff of fric = .4 .4 x normal force = friction force = .4 (44.55kg) = 17.82 kg
to push the box uphill you must overcome the friction force and the downslope force
Pushing force = 17.82 + 22.69 kg = 40.51 kg force
+37084
Correction: I used the wrong coeeficient of friction in my previously posted solution
I used 0.4 instead of the stated 0.1
Here are the corrections:
Friction force = .1 x 44.55 = 4.455
Required pushing force = 4.455kg + 22.69 = 27.15 kg force
Sorry !!!!!
