91Ó°ÊÓ

a) Mass of the slab,M=10 kg

b) Mass of the block,m= 10 kg

c) Force on the slab,F=100 N

To find the acceleration of slab and block, we have to use Newton’s 2nd law of motion. For each given case of slipping and no slipping, we use the coefficient of friction accordingly to determine the minimum and the maximum possible values of the accelerations of the slab and the block.

Formula:

The force according to Newton’s second law,

F = ma (1)

Case (1): No slipping

When no slipping occurs ($\mathrm{μ}$is at its maximum value.), then block and slab stick together.

By using equation (1) along the horizontal direction, we can get the acceleration of the slab as follows:

role="math" localid="1657172766626" $$\begin{gathered} \underset{}{∑}\mathrm{F}=\left(\mathrm{M}+\mathrm{m}\right){\mathrm{a}}_{\mathrm{slab}} \\ {\mathrm{a}}_{\mathrm{slab}}=\frac{\mathrm{F}}{\mathrm{M}+\mathrm{m}} \\ =\frac{100\mathrm{N}}{\left(10\mathrm{mg}+10\mathrm{mg}\right)} \\ =5\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Case (2): In case of slipping

When there is slipping between block and slab ( $\mathrm{μ}=0$), then the slab and block will move separately.

By using equation (1) along the horizontal direction, we get the slab acceleration as follows:

role="math" localid="1657173053653" $$\begin{gathered} \underset{}{∑}\mathrm{F}=M{\mathrm{a}}_{\mathrm{slab}} \\ {\mathrm{a}}_{\mathrm{slab}}=\frac{\mathrm{F}}{\mathrm{M}} \\ =\frac{100\mathrm{N}}{10kg} \\ =10\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore the possible range of acceleration of the slab is$5\mathrm{m}/{\mathrm{s}}^2-10\mathrm{m}/{\mathrm{s}}^2$ .

Case 1: When there is slipping

If the coefficient of friction is 0, i.e.$\mathrm{μ}=0$, then the acceleration of the block is 0 as there is no force acting on the block.

${\mathrm{a}}_{\mathrm{block}}=0\mathrm{m}/{\mathrm{s}}^2$

Case 2: No slipping

If the value of the coefficient of friction is maximum, then block and slab are stick together, then the acceleration of the block is given using equation (1) as:

$$\begin{gathered} \underset{}{∑}\mathrm{F}=\left(\mathrm{M}+\mathrm{m}\right){\mathrm{a}}_{\mathrm{slab}} \\ {\mathrm{a}}_{\mathrm{slab}}=\frac{\mathrm{F}}{\mathrm{M\_m}} \\ =\frac{100\mathrm{N}}{10\mathrm{kg}+10\mathrm{kg}} \\ =5\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore the possible range of acceleration of the block is$0\mathrm{m}/{\mathrm{s}}^2-5\mathrm{m}/{\mathrm{s}}^2$ .