A block with mass $\mathit{m}\mathbf{=}\mathbf{5}\mathbf{.00}\mathbf{\text{\hspace{0.17em}kg}}$slides down a surface inclined $\mathbf{36}\mathbf{.}{\mathbf{9}}^o$to the horizontal (Fig. P10.62). The coefficient of kinetic friction is$\mathbf{0}\mathbf{.}\mathbf{25}$. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass $\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{kg}$and moment of inertia$\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{kg}\mathbf{.}{\mathbf{m}}^2$ with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of $\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{m}$from that axis. (a) What is the acceleration of the block down the plane? (b) What is the tension in the string?

Acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

Calculate the acceleration of the block in the town plane by using net forces act on the block.

The figure represents the free body diagram of all of the forces acting on the block:

Here, the x-axis is define as parallel to the slope with down slope being positive and upslope being negative and y-axis is defined as perpendicular to the slope with positive being away from the hill and negative being in to it. The weight vector $\overrightarrow{w}$is the force on the block due to gravity, ${\overrightarrow{w}}_x$and ${\overrightarrow{w}}_y$are respectively the $x$and $y$component of ${\overrightarrow{w}}_x$$\overrightarrow{N}$is the normal force of the slope of the hill exerts on the block, equal in magnitude to${\overrightarrow{w}}_x$ , $\overrightarrow{f}$is the friction force of the block as it slide down the slope and ${\overrightarrow{f}}_T$is the tension force of the string attached to the pulley

To find the net acceleration of the block down the surface, set up of a forced balance of all forces acting parallel to the slope summering all of the forces acting in the x-direction.

$F_{net,x}\stackrel{⌢}{x}=w_x-fx-F_T\stackrel{⌢}{x}$

Here,$F_{net,x}$is the net force magnitude in the x direction, is the magnitude of x component of the weight, cancel the unit vectors to obtain a force magnitude:

$F_{net,x}=w_x-f-F_T$

To define the tension force of the string between the block and fly wheel, use the definition of the torque magnitude$\tau$produce by a force $F$ acting on a lever arm $I$ from point 0:

$\tau =FI$

The lever arm is just the radius $R$ of the wheel, so

Substitute $F_T$for $F$ and $R$for$I$in $\tau$:

$\tau =F_{T}R$

Set the torque exerted on the wheel$F_{T}R$equal to the definition of torque as the product $I\alpha$of the moment of inertia $I$and the angular acceleration:

$F_{T}R=I\alpha$

The string which connects the block and fly wheel will be pulled down slope by the acceleration block and produce a tangential acceleration on the non-slipping flywheel:

$\alpha =R\alpha$

Rearrange arrange equation for$\alpha$.

$\alpha =\frac{\alpha }{R}$

Substitute $\frac{\alpha }{R}$for$\alpha$ .

$F_{T}R=I\frac{\alpha }{R}$

Rearrange arrange equation for $F_T$.

$F_T=\frac{I}{R^2}\alpha$

The angular of the slope above the horizontal:

$w_s=w\sin \theta$

Total weight is defined as the product

Substitute $mg$ for in the equation $w_s$

$w_s=mg\sin \theta$

Coefficient of kinetic friction ${\mu }_k$ and the magnitude of the normal force$N$:

$f={\mu }_k{w}_f$

Find $w_f$by multiplying the magnitude of the rocks weight$w$by the cosine of angle $\theta$

$w_f=w\cos \theta$

Substitute$mg$ for $w$.

$w_f=mg\cos \theta$

Substitute $mg\cos \theta$ for $w_f$ in $f$.

$f={\mu }_{k}mg\cos \theta$

Substitute $\frac{I\alpha }{R^2}$for $F_T$,$mg\sin \theta$for$w_s$ and

${\mu }_{k}mg\cos \theta$for f in the equation

$f_{net,x}=w_s-f-F_T$

$f_{net,x}=mg\sin \theta -{\mu }_{k}mg\cos \theta -\frac{I}{R^2}\alpha$$f_{net,x}=mg\sin \theta -{\mu }_{k}mg\cos \theta -\frac{I}{R^2}\alpha$

Acceleration of the system a for $f_{net,x}$in the force balance:

$m\alpha =mg\sin \theta -{\mu }_{k}mg\cos \theta -\frac{I}{R^2}\alpha$

Rearrange the equation from a

(a)

$\begin{array}{l}m\alpha +\frac{I}{R^2}\alpha =mg\sin \theta -{\mu }_{k}mg\cos \theta \\ (m+\frac{I}{R^2})\alpha =mg(\sin \theta -{\mu }_k\cos \theta )\\ \alpha =\frac{mg(\sin \theta -{\mu }_k\cos \theta )}{m+\frac{I}{R^2}}\end{array}$

Substitute $5.00\text{\hspace{0.17em}kg}$for m,${\text{36.9}}^{\circ }$for$\theta$ , for $0.25$

For${\mu }_k$,$0.500{\text{\hspace{0.17em}kg.m}}^{{}^2}$for $I$,$0.200\text{\hspace{0.17em}m}$ for $R_{{}_1}$and$9.8{\text{\hspace{0.17em}m/s}}^2$ for g in a.

$\begin{array}{l}\alpha =\frac{\left(5.00\text{\hspace{0.17em}kg}\right)\left(9.8{\text{\hspace{0.17em}m/s}}^2\right)(\sin \left({36.9}^{\circ }\right)-\left(0.25\right)\cos \left({36.9}^{\circ }\right))}{5.00\text{\hspace{0.17em}kg+}\frac{0.500\text{\hspace{0.17em}\hspace{0.17em}kg}.{\text{m}}^2}{{\left(0.200\text{\hspace{0.17em}m}\right)}^2}}\\ =1.12{\text{\hspace{0.17em}m/s}}^2\end{array}$

Therefore, the acceleration of the block down the plane is $1.12{\text{\hspace{0.17em}m/s}}^2$

(b)

The equation for tension in the string is,

$F_T=\frac{1}{R^2}a$

Here, $a$is acceleration of the block.

Substitute $1.12{\text{\hspace{0.17em}m/s}}^2$for a,$0.500\text{\hspace{0.17em}\hspace{0.17em}kg}.{\text{m}}^2$for$I$,

And $0.200\text{\hspace{0.17em}m}$ for R and solve string tension$F_T$.

$\begin{array}{l}\left(\frac{0.500\text{\hspace{0.17em}\hspace{0.17em}kg}.{\text{m}}^2}{{\left(0.200\text{\hspace{0.17em}m}\right)}^2}\right)\left(1.12{\text{\hspace{0.17em}m/s}}^2\right)\\ =14.0\text{\hspace{0.17em}N}\end{array}$

Therefore, the tension in the string is$14.0\text{\hspace{0.17em}N}$.