Question

In Fig. 6-15, a horizontal force of 100Nis to be applied to a 10kg slab that is initially stationary on a frictionless floor, to accelerate the slab. A 10kg block lies on top of the slab; the coefficient of friction $\mathit{\mu }$between the block and the slab is not known, and the block might slip. In fact, the contact between the block and the slab might even be frictionless. (a) Considering that possibility, what is the possible range of values for the magnitude of the slab’s accelerationlocalid="1657173176346" ${\mathbf{a}}_{\mathbf{slab}}$? (Hint:You don’t need written calculations; just consider extreme values for m.) (b) What is the possible range for the magnitudelocalid="1657173167508" ${\mathit{a}}_{\mathbf{b}\mathbf{l}\mathbf{o}\mathbf{c}\mathbf{k}}$of the block’s acceleration?

Short answer

a) The possible range of values for the magnitude of the slab’s acceleration is$5\mathrm{m}/{\mathrm{s}}^2-10\mathrm{m}/{\mathrm{s}}^2$ .

b) The possible range for the magnitude of the block’s acceleration is$0\mathrm{m}/{\mathrm{s}}^2-5\mathrm{m}/{\mathrm{s}}^2$ .

Step-by-step solution

The given data

a) Mass of the slab,M=10 kg

b) Mass of the block,m= 10 kg

c) Force on the slab,F=100 N

Understanding the concept of the friction

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)

a) Calculation of the possible range of slab’s acceleration

Case (1): No slipping

When no slipping occurs ($\mu$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" $$\begin{gathered} \sum _{}\mathrm{F}=\left(\mathrm{M}+\mathrm{m}\right){\mathrm{a}}_{\mathrm{slab}} \\ {\mathrm{a}}_{\mathrm{slab}}=\frac{\mathrm{F}}{\mathrm{M}+\mathrm{m}} \\ =\frac{100\mathrm{N}}{\left(10\mathrm{mg}+10\mathrm{mg}\right)} \\ =5\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Case (2): In case of slipping

When there is slipping between block and slab ( $\mu =0$), 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" $$\begin{gathered} \sum _{}\mathrm{F}=M{\mathrm{a}}_{\mathrm{slab}} \\ {\mathrm{a}}_{\mathrm{slab}}=\frac{\mathrm{F}}{\mathrm{M}} \\ =\frac{100\mathrm{N}}{10kg} \\ =10\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

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

b) Calculation of the possible range of the acceleration of the block

Case 1: When there is slipping

If the coefficient of friction is 0, i.e.$\mu =0$, 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:

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

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