Question

The work done in pulling a body of mass 5 kg along an inclined plane angle 60° with coefficient of friction 0. 2 through 2m will be

Answer 1

Force of friction is down the incline,

f=umgcos@, workdone by friction is Wf =umgcos@ds=-10 as well as one component of weight is acting incline downwards.

Wmg = mgsin@ds=-50sqrt3. Net workdone =change in kinetic energy. Wmg + Wf +W =0.  

W   = -(Wmg + Wf). W =   (10+50sqrt3) and W= 96.5 J

Answer 2

Answer:

Force of friction is down the incline,

f=umgcos@, workdone by friction is Wf =umgcos@ds=-10 as well as one component of weight is acting incline downwards.

Wmg = mgsin@ds=-50sqrt3. Net workdone =change in kinetic energy. Wmg + Wf +W =0.  

W   = -(Wmg + Wf). W =   (10+50sqrt3) and W= 96.5 J