Question

A block of mass $m$ is kept over another block of mass $M$ and the system rests on a horizontal surface. A constant horizontal force $F$ acting on the lower block produces an acceleration $\frac{F}{2(m+M)}$ in the system, the two blocks always move together. (a) Find the coefficient of kinetic friction between the bigger block and the horizontal surface. (b) Find the frictional force acting on the smaller block. (c) Find the work done by the force of friction on the smaller block by the bigger block during a displacement d of the system.

Answer 1

$Giventhatacceleration,a=\frac{F}{2}(M+m)$

$Nowfor\left(a\right):-$

$Weightofthesystem=(M+m)g=N$

$WhereNisthenormalforce.$

$TheforceofFriction=\mu NAlsogiven,$

$a=\frac{F}{2}(m+M)$

$Nowforequationofforce,alongitssurface$

$F-\mu N=(m+M)aF-\mu (m+M)g=(m+M)\frac{F}{2}(M+m)$

$F-\mu (m+M)g=\frac{F}{m}$

$2F-2\mu g(m+M)=F\mu =\frac{F}{2g(m+M)}$

$HencethecoefficientofKineticfrcitionis\frac{F}{2g(m+M)}$

$Nowfor\left(b\right):-$

Frictional force:-Let the frictional force be $f$ acting on the small block so the acceleration must be

$a=\frac{F}{2}(m+M)$

$Sotheforce=m\times a$

$f=m\frac{M}{2}(m+M)f=m\frac{M}{2}(m+M)Hencethefrictionalforceism\frac{F}{2}(m+M).$

$Nowfor\left(c\right):-Workdone:-Thevelocityoftheblockduringdisplacementd,v²=u²+2adv²=2adsinceu=0v²=2F\frac{d}{2}(m+M)v²=Fd(m+M)NowthefinalKineticenergyK.E=\frac{1}{2}\left(mv²\right)K.E=mF\frac{d}{2}(m+M)Sotheworkdoneonsmallerblockislargerthanbiggerblockbythefrictionalforce,sochangeinK.EismF\frac{d}{2}(m+M)$