Chapter 6, Problem 39P

(a)

To determine

Speed of cart at point 3.

Answer to Problem 39P

The speed is $14.3\text{m}/\text{s}$.

Explanation of Solution

Speed of cart at point 1 is $20.0\text{m}/\text{s}$ and $g$ is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Potential energy is measured with respect to point 1.

Potential energy at the level of point 1 is taken as zero. Write the equation for total energy of at point 1.

$E_1=\frac{1}{2}m{v}_1^2$ (I)

Here, $E_1$ is the total energy of cart at point 1, $m$ is the mass of cart, and $v_1$ is the velocity of cart at point 1.

Write the equation for total energy of at point 3.

$E_3=\frac{1}{2}m{v}_3^2+mg{y}_3$ (II)

Here, $E_3$ is the total energy of cart at point 3, $v_3$ is the velocity of cart at point 3, $g$ is the gravitational acceleration, and $y_3$ is the height of point 3 from point 1.

According to the law of conservation of energy, the total energy at point 1 and 3 will remains the same. Equate the right hand sides equations (I) and (II).

$\begin{array}{l}\frac{1}{2}m{v}_1^2=\frac{1}{2}m{v}_3^2+mg{y}_3\\ \frac{1}{2}{v}_1^2=\frac{1}{2}{v}_3^2+g{y}_3\end{array}$

Rewrite the above equation in terms of $v_3$.

$v_3=\sqrt{v_1^2-2g{y}_3}$ (III)

Conclusion:

Substitute $20.0\text{m}/\text{s}$ for $v_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $10.0\text{m}$ for $y_3$ in the above equation to find $v_3$.

$\begin{array}{c}{v}_3=\sqrt{{\left(20.0\text{m}/\text{s}\right)}^2-2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(10.0\text{m}\right)}\\ =\sqrt{204.49}\text{m}/\text{s}\\ =14.3\text{m}/\text{s}\end{array}$

Therefore, the speed is $14.3\text{m}/\text{s}$.

(b)

To determine

Check whether the cart will reach the point 4 or not.

Answer to Problem 39P

Yes. The car will reach point 4.

Explanation of Solution

Speed of cart at point 1 is $20.0\text{m}/\text{s}$ and $g$ is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Rewrite equation (III) by replacing $v_3$ with $v_4$.

$v_4=\sqrt{v_1^2-2g{y}_4}$

Here, $v_4$ is the speed of cart at point 4 and $y_4$ is the height of point 4 from point 1.

Write the condition for the cart to reach 4.

$v_1^2>2g{y}_4$

Conclusion:

Substitute $20.0\text{m}/\text{s}$ for $v_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $20.0\text{m}$ for $y_4$ in the above equation to find whether cart reaches point 4or not.

$\begin{array}{c}{\left(20.0\text{m}/\text{s}\right)}^2>2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(20.0\text{m}\right)\\ 400{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}>392.4{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\end{array}$

Since the condition satisfies, cart can reach at point 4.

Substitute $20.0\text{m}/\text{s}$ for $v_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $20.0\text{m}$ for $y_4$ in the equation for $v_4$.

$\begin{array}{c}{v}_4=\sqrt{{\left(20.0\text{m}/\text{s}\right)}^2-2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(20.0\text{m}\right)}\\ =\sqrt{7.62}\text{m}/\text{s}\\ =2.76\text{m}/\text{s}\end{array}$

Therefore, the cart can reach at point 4.