Consider an object of mass 2 kg moved by an external force 20 N in a surface having coefficient of kinetic friction 0.9 to a distance 10 m. What is the work done by the exteranl force and kinetic friction? Comment on the result. (Assume g = 10 m^(-2))

Consider an object of mass 2 kg moved by an external force 20 N in a s

1.	Consider an object of mass 2 kg moved by an external force 20 N in a surface having coefficient of kinetic friction 0.9 to a distance 10 m. What is the work done by the exteranl force and kinetic friction? Comment on the result. (Assume g = 10 m^(-2))
Answer» SOLUTION :`m=2kg, d=10mF_(ext)=20N=mu_(K)m h=0.9` (a) External force `F_(ext)=20N` (B) Kinetic friction `f_(k)=mu_(k)mg=0.9xx(2)xx10=18N` The work done by the external force `w_(ext)=Fd=20xx10=200J` The work done by the force of kinete friction`w_(k)=f_(k)d=(-18)xx10=-180j` here the negative sin IMPLIES that the force of kinetic friction is opwite to the of displacement The total work done on the object `w_(total)=w_(ext)+w_(k)=200J-180J=20J` Since the friction is a non-conservative force, out of 200J given by external force, the 180J is lost and it can not be recovered.