Question

Question: A block of mass 2.20 kg is accelerated across a rough surface by a light cord passing over a small pulley as shown in Figure P5.99. The tension T in the cord is maintained at 10.0 N, and the pulley is 0.100 m above the top of the block. The coefficient of kinetic friction is 0.400. (a) Determine the acceleration of the block when x = 0.400 m. (b) Describe the general behavior of the acceleration as the block slides from a location where x is large to x= 0 . (c) Find the maximum value of the acceleration and the position x for which it occurs. (d) Find the value of x for which the acceleration is zero.

Short answer

(a) The acceleration of the block when $x=0.4\text{m}$ is $0.933\text{\hspace{0.33em}m}/{\text{s}}^2$.

(b) The acceleration is increases from the value $0.625\text{\hspace{0.33em}m}/{\text{s}}^2$ to the maximum value and then suddenly drops to $-21.0\text{\hspace{0.33em}m}/{\text{s}}^2$ at $x=0$.

(c) The maximum acceleration occurs at $0.250\text{\hspace{0.33em}m}$ and value of maximum acceleration is $0.976\text{\hspace{0.33em}m}/{\text{s}}^2$.

(d) The value of the x at which acceleration is zero is $6.10\text{\hspace{0.33em}cm}$.

Step-by-step solution

Writing the given data from the question

Consider the given data as below.

The mass of the block, $m=2.20\text{\hspace{0.33em}kg}$

Coefficient of kinetic friction, a

The tension in the cord, T= 10 N

The distance from the above the block, h= 0.100 m

Acceleration due to gravity, g= 9.8 m/s²

Determining the required formulas

The expression to calculate the force on the object is given as follows.

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$

Here, F is the working acting on object, m is the mass of the object and, a is the acceleration of the object.

The expression to calculate the frictional force is given as follows.

${\mathit{f}}_{\mathbf{s}}\mathbf{=}{\mathit{\mu }}_{\mathbf{s}}\mathit{n}$ ……. (i)

Here, n is the normal force acting on the block.

(a) Calculating the acceleration of the block when  

Consider the figure shown below which shows the accelerated block through the rough surface.

Calculate the angle of the cord.

$\tan \theta =\frac{h}{x}$

Substitute 0.1 m for h and 0.400 m for x into above equation.

$$\begin{gathered} \begin{array}{c}\tan \theta =\frac{0.1}{0.4}\\ =0.25\end{array} \\ \begin{array}{c}\theta ={\tan }^{-1}\left(0.25\right)\\ =14°\end{array} \end{gathered}$$

Let assume n is the normal force acting on the block.

Consider the equilibrium along the y - axis.

$\begin{array}{l}\sum F_y=m{a}_y\\ T\sin \theta -mg+n=0\end{array}$

Substitute 10 N for T , 14^o for θ , 2.2 kg form and 9.8 m/s² forg into above equation.

$\begin{array}{c}10\times \sin 14°-2.2\times 9.8+n=0\\ 2.419-21.56+n=0\\ -19.141+n=0\\ n=19.141\text{N}\end{array}$

Calculate the frictional force,

Substitute 0.400 for ${\mu }_s$ and $19.141\text{\hspace{0.33em}N}$ for n into equation (i).

$\begin{array}{c}{f}_s=0.4\times 19.141\\ =7.65\text{\hspace{0.33em}N}\end{array}$

Consider the equilibrium along thex - axis.

$\begin{array}{l}\sum F_x=ma\\ T\cos \theta -f_s=ma\end{array}$

Substitute 10N for T , 14^o for θ , 2.2 kg for m and 7.65 N for f_s into the above equation.

$\begin{array}{c}10\cos 14°-7.65=2.2\times a\\ 9.702-7.65=2.2a\\ 2.052=2.2a\\ a=0.933\text{\hspace{0.33em}m}/{\text{s}}^2\end{array}$

Hence, the acceleration of the block when x= 0.4 m is 0.933 m/s².

(b) Describing the general behaviour of the block when block slides from the large value to zero

When the value of x is large then weight is equal to normal reaction.

$n=mg$

Substitute 2.2 kg for m and 9.8 m/s² for g into above equation.

$\begin{array}{c}n=2.2\times 9.8\\ =21.56\text{\hspace{0.33em}N}\end{array}$

Calculate the frictional force by substitute 0.4 for ${\mu }_s$ and 21.56 N for n into equation (i).

$\begin{array}{c}{f}_s=0.4\times 21.56\\ =8.624\text{N}\end{array}$

The acceleration is given by

$\begin{array}{l}ma=T-f_s\\ a=\frac{T-f_s}{m}\end{array}$

Substitute 10 N for T , 2.2 kg for m and 8.624 N for f_s into above equation.

$\begin{array}{c}a=\frac{10-8.624}{2.2}\\ =\frac{1.376}{2.2}\\ =0.625\text{\hspace{0.33em}m}/{\text{s}}^2\end{array}$

When x = 0, then the acceleration of the block is given by

$\begin{array}{c}{a}_{x=0}=\frac{0-{\mu }_s\left(n-T\right)}{m}\\ =-\frac{{\mu }_s\left(n-T\right)}{m}\end{array}$

Substitute 10 N for T ,2.2 kg for m , 21.56 N for n , and 0.4 for ${\mu }_s$ into above equation.

$\begin{array}{c}{a}_{x=0}=-\frac{0.4\left(21.56\text{\hspace{0.33em}N}-10\text{\hspace{0.33em}N}\right)}{2.2\text{\hspace{0.33em}kg}}\\ =-\frac{4.624}{2.2}\\ =-2.10\text{\hspace{0.33em}m}/{\text{s}}^2\end{array}$

Therefore, from the above discussion, it can be concluded that the acceleration is increases from the value 0.625 m/s² to the maximum value and then suddenly drops to $-21.0\text{\hspace{0.33em}m}/{\text{s}}^2$at x= 0.

(c) Calculating the maximum value of acceleration and its location

The acceleration of the block is given by

$$\begin{gathered} ma=T\cos \theta -{\mu }_s\left(n-T\sin \theta \right) \\ a=\frac{T\cos \theta -{\mu }_s\left(n-T\sin \theta \right)}{m}............................\left(ii\right) \end{gathered}$$

Calculate the value of $\cos \theta$from the diagram.

$\begin{array}{c}\cos \theta =\frac{x}{\sqrt{x^2+h^2}}\\ =\frac{x}{\sqrt{x^2+{0.1}^2}}\end{array}$

Calculate the value of $\sin \theta$from the diagram.

$\begin{array}{c}\sin \theta =\frac{h}{\sqrt{x^2+h^2}}\\ =\frac{0.1}{\sqrt{x^2+{0.1}^2}}\end{array}$

Substitute 10 N for T, $\frac{x}{\sqrt{x^2+{0.10}^2}}$ for $\cos \theta$, $\frac{0.1}{\sqrt{x^2+{0.1}^2}}$ for $\sin \theta$ , 21.56 N for n, 0.4 for ${\mu }_s$ and 2.2 kg for m into equation (ii).

$a=\frac{\left(10\text{\hspace{0.33em}N}\right)\frac{x}{\sqrt{x^2+{0.1}^2}}-0.4\left(21.56\text{\hspace{0.33em}N}-10\text{N}\frac{0.1}{\sqrt{x^2+{0.1}^2}}\right)}{2.2\text{\hspace{0.33em}kg}}$

$a=4.55\frac{x}{\sqrt{x^2+{0.1}^2}}-3.92+0.182\frac{1}{\sqrt{x^2+{0.1}^2}}......................\left(iii\right)$

Differentiate the above equation with respect to x.

$\begin{array}{c}\frac{da}{dx}=\left\{4.55{\left(x^2+{0.1}^2\right)}^{-\frac{1}{2}}+4.55\times x\left(-\frac{1}{2}\right)\left(2x\right){\left(x^2+{0.1}^2\right)}^{-\frac{3}{2}}+0.182\left(-\frac{1}{2}\right)\left(2x\right){\left(x^2+{0.1}^2\right)}^{-\frac{3}{2}}\right\}\\ =4.55{\left(x^2+{0.1}^2\right)}^{-\frac{1}{2}}-4.55x^2{\left(x^2+{0.1}^2\right)}^{-\frac{3}{2}}-0.182x{\left(x^2+{0.1}^2\right)}^{-\frac{3}{2}}\end{array}$

Substitute 0 for$\frac{da}{dx}$into above equation.

$\begin{array}{l}\frac{da}{dx}=0\\ 4.55{\left(x^2+{0.1}^2\right)}^{-\frac{1}{2}}-4.55x^2{\left(x^2+{0.1}^2\right)}^{-\frac{3}{2}}-0.182x{\left(x^2+{0.1}^2\right)}^{-\frac{3}{2}}=0\\ {\left(x^2+{0.1}^2\right)}^{-\frac{3}{2}}\left[4.55\left(x^2+{0.1}^2\right)-4.55x^2-0.182x\right]=0\\ 4.55\left(x^2+{0.1}^2\right)-4.55x^2-0.182x=0\end{array}$

Solve further as

$\begin{array}{l}4.55x^2+0.0455-4.55x^2-0.182x=0\\ 0.0455-0.182x=0\\ 0.182x=0.0455\\ x=250\text{\hspace{0.33em}m}\end{array}$

Substitute 0.250 m for x into equation (iii).

$$\begin{gathered} a=4.55\left(\frac{0.250}{\sqrt{{0.250}^2+{0.1}^2}}\right)-3.92+0.182\left(\frac{1}{\sqrt{{0.250}^2+{0.1}^2}}\right) \\ \begin{array}{c}a=\frac{1.1375}{0.269}-3.92+0.182\left(\frac{1}{0.269}\right)\\ =4.228-3.92+0.676\\ =0.976\text{m}/{\text{s}}^2\end{array} \end{gathered}$$

Hence, the maximum acceleration occurs at 0.250 m and value of maximum acceleration is 0.986 m/s².

(d) Calculating the location at point where the acceleration is zero

Calculate the location where acceleration is zero.

Substitute 0 for a into equation (iii).

$\begin{array}{l}0=4.55\frac{x}{\sqrt{x^2+{0.1}^2}}-3.92+0.182\frac{1}{\sqrt{x^2+{0.1}^2}}\\ 3.92=4.55\frac{x}{\sqrt{x^2+{0.1}^2}}+\frac{0.182}{\sqrt{x^2+{0.1}^2}}\\ 3.92\sqrt{x^2+{0.1}^2}=4.55x+0.182\end{array}$

Take square of both the sides of the equation.

$\begin{array}{l}{\left(3.92\sqrt{x^2+{0.1}^2}\right)}^2={\left(4.55x+0.182\right)}^2\\ 15.4\left(x^2+0.01\right)=20.7x^2+0.0331+1.65x\\ 15.4x^2+0.154=20.7x^2+0.0331+1.65x\\ 5.3x^2+1.65x-0.121=0\end{array}$

Solve the quadratic equation

$\begin{array}{c}x=\frac{-1.65\pm \sqrt{{\left(1.65\right)}^2-4\left(5.3\right)\left(-0.121\right)}}{2\left(5.3\right)}\\ =\frac{-1.65\pm \sqrt{2.7225+2.5652}}{10.6}\\ =\frac{-1.65\pm \sqrt{5.2877}}{10.6}\\ =\frac{-1.65\pm 2.2995}{10.6}\end{array}$

$\begin{array}{c}x=\frac{-1.65+2.2995}{10.6}\\ =0.0610\text{m}\\ =6.\text{10cm}\end{array}$

Hence, the value of the x at which acceleration is zero is 6.10 cm.