Question

A block of mass ‘m’ is placed over another block of mass M. Lower block is connected with a spring of force constant ‘k’ as shown. There is no friction between the lower block and ground, and friction coefficient between both blocks is μ. If blocks are displaced to the right by x and then released, then maximum value of x, so that both blocks move together without any slipping between them, is:

• A. μg(m + M)/k
• B. μg(m + M)/2k
• C. 2μg(m + M)/2k
• D. μg(2m + M)/k

Answer 1

Answer: (A)

μg(m + M)/k

Answer 2

If both blocks have to move together then there must not be any slipping at the moment of release as acceleration is maximum at this point. Hence maximum friction force will be required to act at this moment that must be less than or equal to limiting value so that there is not slipping between two blocks.

a_M = a_m(maximum) for x to be maximum

Maximum acceleration of block of mass ‘m’ = μmg/m = μg

act at this moment that must be less than or equal to limiting value so that there is no slipping between two blocks.. $a _ { M } = a _ { m ( \text { maximum) } }$ for $x$ to be maximum Maximum acceleration of block of mass ' $m ^ { \prime } = \frac { \mu \mathrm { mg } } { \mathrm { m } } = \mu \mathrm { g }$ thus for lower block $\frac { \mathrm { kx } _ { \max } - \mu \mathrm { mg } } { \mathrm { M } } = \mu \mathrm { g }$ $\mathrm { x } _ { \max } = \frac { \mu \mathrm { Mg } + \mu \mathrm { mg } } { \mathrm { k } } = \frac { \mu \mathrm { g } ( \mathrm { m } + \mathrm { M } ) } { \mathrm { k } }$

thus for lower block

act at this moment that must be less than or equal to limiting value so that there is no slipping between two blocks.. $a _ { M } = a _ { m ( \text { maximum) } }$ for $x$ to be maximum Maximum acceleration of block of mass ' $m ^ { \prime } = \frac { \mu \mathrm { mg } } { \mathrm { m } } = \mu \mathrm { g }$ thus for lower block $\frac { \mathrm { kx } _ { \max } - \mu \mathrm { mg } } { \mathrm { M } } = \mu \mathrm { g }$ $\mathrm { x } _ { \max } = \frac { \mu \mathrm { Mg } + \mu \mathrm { mg } } { \mathrm { k } } = \frac { \mu \mathrm { g } ( \mathrm { m } + \mathrm { M } ) } { \mathrm { k } }$

$\frac { \mathrm { kx } _ { \max } - \mu \mathrm { mg } } { \mathrm { M } } = \mu \mathrm { g }$

x_max = $\frac { \mu \mathrm { Mg } - \mu \mathrm { mg } } { \mathrm { k } } = \frac { \mu \mathrm { g } ( \mathrm { m } + \mathrm { M } ) } { \mathrm { k } }$