Figure 6-20 shows an initially stationary block of masson a floor. A force of magnitudeis then applied at upward angle$\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{20}\mathbf{°}$.What is the magnitude of the acceleration of the block across the floor if the friction coefficients are (a)$\mathit{\mu }\mathit{s}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{600}$and${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{500}$and (b)$\mathit{\mu }\mathit{s}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{400}$and${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{300}$?

Mass of the block, m

Force on the block,$F=0.500mg.N$

Angle,$\theta ={20}^0$

${\mu }_s=$0.600 and ${\mu }_K=$ 0.500

${\mu }_s=$0.400 and ${\mu }_K=$0.300

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motionalong vertical and horizontal direction.

Write the formula for the net force:

$F_{net}=\sum ma$

Here, F is the net force, m is mass and a is an acceleration.

By using Newton’s 2nd law along y direction,block is not moving along y. So, the acceleration is 0.

$\sum _{}^{}{F}_y=0$

$N+F\sin \left(20\right)-mg=0$

Substitute the values and solve as:

$$\begin{gathered} N=mg-\left(0.500\text{mg}\right)\sin \left(20\right) \\ =0.83mgN \end{gathered}$$

Static frictional force on the block from the floor is,

$\begin{array}{rcl}{f}_s& =& {\mu }_{s}N\\ & =& \left(0.600\right)\left(0.83\text{mg}\right)\\ & =& 0.50\text{mgN}\\ & & \end{array}$

Opposite force to the$f_s$is,

$\begin{array}{rcl}F\cos \left(20\right)& =& \left(0.500\text{mg}\right)\cos \left(20\right)\\ & =& 0.47\text{mg}·\text{N}\\ & & \end{array}$

So, here,$F\cos \left(20\right)<f_s$therefore block cannot move.

Therefore, the magnitude of the acceleration of the block across the floor is 0 .$\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Static frictional force on the block from the floor is,

$\begin{array}{rcl}{f}_s& =& {\mu }_{s}N\\ & =& \left(0.400\right)\left(0.83\text{mg}\right)\\ & =& 0.33\text{mg}·\text{N}\\ & & \end{array}$

Opposite force to the$f_s$is,

$\begin{array}{rcl}F\cos \left(20\right)& =& \left(0.500mg\right)\cos \left(20\right)\\ & =& 0.47mg.N\end{array}$

So,here. $F\cos \left(20\right)>f_s$Therefore, the block will be moved.

Kinetic frictional force on the block is,

$\begin{array}{rcl}{f}_k& =& {\mu }_{k}N\\ & =& \left(0.300\right)\left(0.83\text{mg}\right)\\ & =& 0.25mg.N\end{array}$

By using Newton’s 2nd law along x direction is,

$$\begin{gathered} \sum _{}^{}{F}_x=m{a}_x \\ F\cos \left(20\right)-f_k=m{a}_x \\ \left(0.500mg\right)\cos \left(20\right)-\left(0.25mg\right)=m{a}_x \\ {a}_x=2.15\frac{\text{m}}{{\text{s}}^{\text{2}}} \end{gathered}$$

Therefore, the magnitude of the acceleration of the block across the floor is 2.15.$\frac{\text{m}}{{\text{s}}^{\text{2}}}$