Work being a scalar quantity sometimes has negative and positive signs. What does the sign suggest?
a) The direction of work
b) Relative directions of force and displacement
c) The direction of force
d) The direction of displacement

Explanation

a. Take the 2 kg block as the system. The forces on this block are shown in ure with `M=2` kg. It is assumed that m has its minimum value so that the 2 kg block has a tendency to slip down.As the block is in equilibrium, the resutant force shoud be zero.


Taking coponents `_|_` to the incline
`N=Mg cos 45^0=Mg /sqrt2`.
Taking components || to the incline
`T+f=Mg sin 45^0 = Mg /sqrt2`. or ` T=Mgsqrt2-f`
As it is a case of limiting equilibrium.
`f=mu_2N`
or `T=Mg/sqrt2-mu_s Mg/sqrt2=Mg/sqrt(1-mu_2)`............i
Now consider the other block as the system. the forces acting on this block are shown in ure.
Taking components `_|_` to the incline,
`N'=mg cos 45^0+f'=mg/sqrt2+f'`
As it is the case of limiting equilirium
`f'=mu_sN'=mu_s mg/sqrt2`
Thus, `T=mg/sqrt2(1+mu_s)`
From i. and ii
`m(1+mu_s)=M(1-mu_s)`
or `m=(1-mu_s)/(1+mu_s) M=(1-0.28)/(1+0.28)xx 2 kg`
when maximum possible value of m is suplied the directions of friction are reversed becauyse m hs the tendency to sip down and 2 kg block to slip up. Thus, themaximum value of m can be obtained from iii by utting `mu=-0.28. Thus the maximum value of m is `brgt `m=(1+0.28)/(1-0.28)xx2 kg
`32/9kg`
b. If `m=9/8 kg ` and the system is gently pushed, kinetic friction will operate. Thus,
`f=mu_k Mg/sqrt2 and f'(mu_kmg)/sqrt2
where `mu_k=0.25`. If the acceleration is a Newton's second law for M gives ure.
`Mg sin 45^0-T-f=Ma`
or, `Mg/sqrt2-T-(mu_kMg)/sqrt2Ma`............iv
Applying Newton'ssecond law m ure.
`T-mg sin 45^0 -f' =Ma`
`T-Mg/sqrt2-(mu_kmkg)/sqrt2=Ma`a...........v
Adding iv and v
`Mg/sqrt2(1-mu_k)mg/sqrt2(1+mu_k)=(M+m)a`
or, `a=(M(1-mu_k)-m(1+mu_k))/(sqrt(2(M+m)) g`
`=(2xx0.75-9/8xx1.25)/(sqrt2(2+9/8))g`
`=031 m/s^2`.