Chapter 6, Problem 97QAP

To determine

(a)

The distance up to which the second ramp object slide before coming to a momentary stop.

Answer to Problem 97QAP

The distance up to which the second ramp object slide before coming to a momentary stop is $19.94\text{m}$.

Explanation of Solution

Given:

The given figure is,

Figure 1

The values are:

  $\begin{array}{l}{H}_1=12.0\text{m}\\ H_2=7.00\text{m}\\ {\theta }_1=60.0°\\ {\theta }_2=37.0°\end{array}$

Concept Used:

Interpreting kinetic energy.

Calculation:

Let us consider the kinetic energy at the starting point:

  $\begin{array}{l}{K}_i=\frac{1}{2}m{v}_i^2=0\\ U_{grav,i}=mg{y}_i=mg{H}_1\end{array}$

At the point where the second ramp ends kinetic energy is:

  $\begin{array}{l}{K}_f=\frac{1}{2}m{v}_f^2\\ U_{grav,f}=mg{y}_f=mgd\sin {\theta }_2\end{array}$

According to the law of conservation of energy:

  $\begin{array}{l}{U}_i+K_i=U_f+K_f\\ 0+mg{H}_1=mgd\sin {\theta }_2+0\\ d=\frac{H_1}{\sin {\theta }_2}\end{array}$

On replacing the values, we get

  $\begin{array}{l}{H}_1=12.0\text{m}\\ {\theta }_2=37.0°\\ d=\frac{12.0}{\sin 37.0}\\ d=19.94\text{m}\end{array}$

Conclusion:

Thus, the distance up to which the second ramp object slide before coming to a momentary stop is $19.94\text{m}$.

To determine

(b)

The speed of the object at the height of $H_2=7.00\text{m}$.

Answer to Problem 97QAP

The speed of the object at the height of $H_2=7.00\text{m}$ is $9.90\text{m/sec}$.

Explanation of Solution

Given:

The given figure is,

Figure 1

The values are:

  $\begin{array}{l}{H}_1=12.0\text{m}\\ H_2=7.00\text{m}\\ {\theta }_1=60.0°\\ {\theta }_2=37.0°\end{array}$

Concept Used:

Interpreting potential energy.

Calculation:

Let us consider the kinetic energy at the starting point:

  $\begin{array}{l}{K}_i=\frac{1}{2}m{v}_i^2=0\\ U_{grav,i}=mg{y}_i=mg{H}_1\end{array}$

At the point where the second ramp ends kinetic energy is:

  $\begin{array}{l}{K}_f=\frac{1}{2}m{v}_f^2\\ U_{grav,f}=mg{y}_f=mg{H}_2\end{array}$

According to the law of conservation of energy:

  $\begin{array}{l}{U}_i+K_i=U_f+K_f\\ 0+mg{H}_1=mg{H}_2+\frac{1}{2}m{v}_f^2\\ v_f=\sqrt{2g(H_1-H_2)}\end{array}$

On replacing the values, we get

  $\begin{array}{l}{H}_1=12.0\text{m}\\ H_2=7.00\text{m}\\ g=9.8{\text{m/sec}}^{\text{2}}\\ v_f=\sqrt{2\times 9.8(12.0-7.00)}\\ v_f=9.90\text{m/sec}\end{array}$

Conclusion:

The speed of the object at the height of $H_2=7.00\text{m}$ is $9.90\text{m/sec}$.