Question

a body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4 respectively.What is the acceleration on applying the same force?

Answer 1

Explanation:

See the maximum static friction that can act on the body is Umg = 0.5*60*9.8 = 294 N

now since the force is just above 300 N, the body will translate and the type of friction now will be kinetic instead of static. Hence the net acceleration is 294-0.4*60*9.8/60 = 0.98 m/s^2 Ans.

Answer 2

Explanation:

maximum static friction is 0.5*60*9.8=294N

maximum static friction is 0.5*60*9.8=294Nmax kinetic friction is 0.4*60*9.8=235.2N

maximum static friction is 0.5*60*9.8=294Nmax kinetic friction is 0.4*60*9.8=235.2Nso,avg force is 294 - 235.2 = 58.8N

maximum static friction is 0.5*60*9.8=294Nmax kinetic friction is 0.4*60*9.8=235.2Nso,avg force is 294 - 235.2 = 58.8Nhence F = ma

maximum static friction is 0.5*60*9.8=294Nmax kinetic friction is 0.4*60*9.8=235.2Nso,avg force is 294 - 235.2 = 58.8Nhence F = ma a = F/m

maximum static friction is 0.5*60*9.8=294Nmax kinetic friction is 0.4*60*9.8=235.2Nso,avg force is 294 - 235.2 = 58.8Nhence F = ma a = F/m a = 58.8/60

maximum static friction is 0.5*60*9.8=294Nmax kinetic friction is 0.4*60*9.8=235.2Nso,avg force is 294 - 235.2 = 58.8Nhence F = ma a = F/m a = 58.8/60 = 0.98m/s^2