Chapter 4, Problem 167P

(a)

To determine

The magnitude and direction of the acceleration of pumpkin and the watermelon.

Answer to Problem 167P

The magnitude of acceleration of pumpkin and the watermelon is $0.34{\text{m/s}}^2$ the direction of watermelon slides up the ramp and pumpkin slides down.

Explanation of Solution

The figure 1 shows net force exerted on the watermelon and pumpkin attached by the cord

Since the watermelon and pumpkin are attached by the cord, they must have the same magnitude of acceleration.

Write the equation of the net force exerted on the pumpkin by using Newton’s second law.

$\begin{array}{c}{\displaystyle \sum F_x}=m_{\text{p}}a\\ m_{\text{p}}g\sin {\theta }_{\text{p}}-T=m_{\text{p}}a\end{array}$ (I)

Here, $m_{\text{p}}$ is the mass of the pumpkin, $g$ is the gravitational acceleration, $T$ is the tension exerted by the force on pumpkin attached by the cord, ${\theta }_{\text{p}}$ is the angle between the pumpkins from the normal line, and $a$ is the acceleration of the pumpkin.

Write the equation of the net force exerted on the watermelon by using Newton’s second law.

$\begin{array}{c}{\displaystyle \sum F_x}=m_{\text{w}}a\\ T-m_{\text{w}}g\sin {\theta }_{\text{w}}=m_{\text{w}}a\end{array}$ (II)

Here, $m_{\text{w}}$ is the mass of the watermelon, $T$ is the tension exerted by the force on watermelon attached by the cord, ${\theta }_{\text{w}}$ is the angle between the watermelons from the normal line, and $a$ is the acceleration of the watermelon.

Solve and add the equation (I) and (II) for acceleration.

$\begin{array}{c}\underset{\_}{\begin{array}{l}{m}_{\text{p}}g{\sin }_{\text{p}}\theta -T=m_{\text{p}}a\\ T-m_{\text{w}}g\sin {\theta }_{\text{w}}=m_{\text{w}}a\end{array}}\\ g\left(m_{\text{p}}\sin {\theta }_{\text{p}}-m_{\text{w}}\sin {\theta }_{\text{w}}\right)=\left(m_{\text{p}}+m_{\text{w}}\right)a\\ a=\frac{g\left(m_{\text{p}}\sin {\theta }_{\text{p}}-m_{\text{w}}\sin {\theta }_{\text{w}}\right)}{\left(m_{\text{p}}+m_{\text{w}}\right)}\end{array}$

Conclusion:

Substitute $9.80{\text{m/s}}^2$ for $g$, $7.00\text{kg}$ for $m_{\text{p}}$, $53.0°$ for ${\theta }_{\text{p}}$, $10.0\text{kg}$ for $m_{\text{w}}$, and $30.0°$ for ${\theta }_{\text{w}}$ in above equation.

$\begin{array}{c}a=\frac{\left(9.80{\text{m/s}}^2\right)\left[\left(7.00\text{kg}\right)\sin \left(53.0°\right)-\left(10.0\text{kg}\right)\sin \left(30.0°\right)\right]}{\left(7.00\text{kg}+10.0\text{kg}\right)}\\ =\frac{5.786\text{kg}\cdot {\text{m/s}}^2}{17.0\text{kg}}\\ =0.34{\text{m/s}}^2\end{array}$

The acceleration is positive, so the direction of watermelon slides up the ramp and pumpkin slides down.

Therefore, the magnitude of acceleration of pumpkin and the watermelon is $0.34{\text{m/s}}^2$. Here the watermelon moves up and to the left.

(b)

To determine

The distance of the pumpkin after release from the inclined path.

Answer to Problem 167P

The distance of the pumpkin after release from the inclined path is $1.5\text{cm}$.

Explanation of Solution

Write the equation of motion for the pumpkin travel down the ramp.

$\Delta x=u\Delta t+\frac{1}{2}a\Delta t^2$ (III)

Here, $\Delta x$ is the total distance of the pumpkin slides from the ramp, $u$ is the initial velocity of the pumpkin, $a$ is acceleration of pumpkin, and $\Delta t$ is the traveling time of the pumpkin.

Conclusion:

Substitute $0\text{m/s}$ for $u$, $0.30\text{s}$ for $\Delta t$, and $0.34{\text{m/s}}^2$ for $a$ in equation (III).

$\begin{array}{c}\Delta x=\left(0\text{m/s}\right)\left(0.30\text{s}\right)+\frac{1}{2}\left(0.34{\text{m/s}}^2\right){\left(0.30\text{s}\right)}^2\\ =0.015\text{m}\\ =0.015\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =1.5\text{cm}\end{array}$

Therefore, the distance of the pumpkin after release from the inclined path is $1.5\text{cm}$.

(c)

To determine

The speed of the watermelon.

Answer to Problem 167P

The speed of the watermelon is $6.8\text{cm/s}$.

Explanation of Solution

Write the equation of motion for the watermelon gains its speed down the ramp.

$v=u+a\Delta t$ (IV)

Here, $v$ is the speed of the watermelon, $u$ is the initial velocity of the watermelon, $a$ is acceleration of watermelon, and $\Delta t$ is the traveling time of the watermelon.

Conclusion:

Substitute $0\text{m/s}$ for $u$, $0.20\text{s}$ for $\Delta t$, and $0.34{\text{m/s}}^2$ for $a$ in equation (IV).

$\begin{array}{c}v=\left(0\text{m/s}\right)+\left(0.34{\text{m/s}}^2\right)\left(0.20\text{s}\right)\\ =0.068\text{m/s}\\ =0.068\text{m/s}\left(\frac{100\text{cm/s}}{1\text{m/s}}\right)\\ =6.8\text{cm/s}\end{array}$

Therefore, the speed of the watermelon is $6.8\text{cm/s}$.