Question

1.A 65.0 kg object slides down a 30^o angle slope from rest at a height of 15.0 m. The coefficient of kinetic friction between the object and the slope surface is 0.15. (A) What is the object

Answer 1

If this were a frictionless slide, the conservation of energy law (the physics invoked here) dictates that KE = 1/2 mV^2 = mgH = PE, which means all the potential enrgy at H = 5 m height is converted into kinetic energy by the time the object reaches the bottom of the slope. In which case V = sqrt(2gH) would be true for the end speed.

But, this is important, this is not frictionless. Friction saps the energy and casts it off as heat; and that lost enregy is not converted iinto kinetic energy. So we really have KE + QE = PE, where QE = kN = kmg cos(theta) is the friction energy lost and k = .15 and theta = 30 deg. N is the normal weight of the object. This revised equation, to refelct friction, says the potential energy will be converted in some part into KE and in some part into QE.

Thus 1/2 mv^2 + kmg cos(theta) = mgH and solve for v = sqrt(2g(H - k cos(theta))) = sqrt(2*9.81*(5 - .15*cos(30))) = 9.78 mps ANS.

Note the physics. This result applies to any mass size as m, the mass, cancels out. Also note that as V^2 = 2gH for the frictionless case, v = sqrt(2gH - 2gk cos(theta)) = sqrt(V^2 - 2gk cos(theta)) which clearly shows that v < V whenever there is friction. In other words, the physics you should learn, is that frcition always decrements the end speed.

F= mg sin 30 -u mg cos30 = ma
a= g(sin30 - u cos30)
v^2 = 2a x
v =[2g(sin30 - u cos30) * x ]^.5 Ans
u=.15 x=5