Chapter 6, Problem 142P

(a)

To determine

The speed of block by considering the energy.

Answer to Problem 142P

The speed of block is $2.0\text{m/s}$.

Explanation of Solution

According to work energy theorem, the work performed by a system will be equivalent to the change in its kinetic energy.

Write an expression for work energy theorem for the block.

$\begin{array}{c}{W}_{total}=\Delta K\\ W_{grav}+W_{friction}=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2\\ mgd\sin \theta -{\mu }_{k}mgd\cos \theta =\frac{1}{2}m{v}_f^2-\frac{1}{2}m{\left(0\right)}^2\\ mgd\sin \theta -{\mu }_{k}mgd\cos \theta =\frac{1}{2}m{v}_f^2\end{array}$ (I)

Here, $W_{total}$ is the total work done, $\Delta K$ is the change in kinetic energy, $W_{grav}$ is the work done by gravity, $W_{friction}$ is the work done by the frictional force, $m$ is the mass of the block, $g$ is the acceleration due to gravity, $v_f$ is the final velocity of the block, $v_i$ is the initial velocity of the block, $d$ is the displacement of the block, ${\mu }_k$ is the coefficient of kinetic friction, and the inclination is $\theta$.

Rewrite the expression (I) to calculate $v_f$.

$v_f=\sqrt{2gd\left(\sin \theta -2{\mu }_k\cos \theta \right)}$ (II)

Conclusion:

Substitute $9.80{\text{m/s}}^2$ for $g$, $0.38$ for ${\mu }_k$, $60.0°$ for $\theta$ and $30.0\text{cm}$ for $d$ in equation (II) to find $v_f$.

$\begin{array}{c}{v}_f=\sqrt{2\left(9.80{\text{m/s}}^2\right)\left(\left(30.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)\left(\sin \left(60.0°\right)-2\left(0.38\right)\cos \left(60.0°\right)\right)}\\ =\sqrt{2\left(9.80{\text{m/s}}^2\right)\left(30.0\times {10}^{-2}\text{m}\right)\left(\sin \left(60.0°\right)-2\left(0.38\right)\cos \left(60.0°\right)\right)}\\ =2.0\text{m/s}\end{array}$

Thus, the speed of block is $2.0\text{m/s}$.

(b)

To determine

The speed of block using Newton’s second law.

Answer to Problem 142P

The speed of block is $2.0\text{m/s}$.

Explanation of Solution

According to Newton’s second law of motion, the net force acting on a system is the product of mass of the system and the acceleration of the system.

Write an expression for Newton’s second law of motion.

$\begin{array}{c}\Sigma F_x=ma\\ mg\sin \theta -{\mu }_{k}mg\cos \theta =ma\\ g\sin \theta -{\mu }_{k}g\cos \theta =a\\ g\left(\sin \theta -{\mu }_k\cos \theta \right)=a\end{array}$ (III)

Here, $\Sigma F_x$ is the net horizontal force on the block and $a$ is the acceleration of the block.

Write an expression to calculate the final velocity of the block.

$\begin{array}{c}{v}_f=\sqrt{v_i^2+2ad}\\ =\sqrt{\left(0^2\right)+2\left(g\left(\sin \theta -{\mu }_k\cos \theta \right)\right)d}\\ =\sqrt{2gd\left(\sin \theta -2{\mu }_k\cos \theta \right)}\end{array}$ (IV)

Conclusion:

Substitute $9.80{\text{m/s}}^2$ for $g$, $0.38$ for ${\mu }_k$, $60.0°$ for $\theta$ and $30.0\text{cm}$ for $d$ in equation (IV) to find $v_f$.

$\begin{array}{c}{v}_f=\sqrt{2\left(9.80{\text{m/s}}^2\right)\left(\left(30.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)\left(\sin \left(60.0°\right)-2\left(0.38\right)\cos \left(60.0°\right)\right)}\\ =\sqrt{2\left(9.80{\text{m/s}}^2\right)\left(30.0\times {10}^{-2}\text{m}\right)\left(\sin \left(60.0°\right)-2\left(0.38\right)\cos \left(60.0°\right)\right)}\\ =2.0\text{m/s}\end{array}$

Thus, the speed of block is $2.0\text{m/s}$.