A force of 40 N acts on body of mass 5 Kg. What is the amount of work done if the total displacement is 2 m and is in the direction of the force applied?
a) 80 J
b) -80 J
c) 40 J
d) 16 J

Explanation

ure shows the forces acting on the two blocks. As we aree looking for the maximum value off M/m. the equilibrium is limiting. Hence, the frictioN/Al forces are equal to `mu` times the corresponding normal forces.

elquilibrium of the block m gives
T=muN_1 and N_1=mg`
which gives
`T=mu mg`......i
Next, consider the equilibrium of the block M. Taking components parallel to the incline
T+muN_2=Mg sintheta`
Taking components normal to the incline
`N_2=Mg costheta`
These give `T=Mg(sintheta-mu costheta)`..........ii
From i and i `mu mg=Mg(sintheta-mu cotheta)`
or, `M/m=mu/(sintheta-mucosthheta)`
`tanthetaltmu, (sinthet-mu costheta)lt0` and the system wil not slide for any value of M/m.