Chapter 6, Problem 11E

(a)

To determine

The increase in potential energy of the rock.

Answer to Problem 11E

The increase in potential energy of the rock is $88.2\text{J}$.

Explanation of Solution

Given info: The mass of the rock is $5.0\text{kg}$ and the height by which the rock is lifted is $1.8\text{m}$.

Write the expression for the increase in potential energy.

$\Delta PE=P{E}_{\text{height}}-P{E}_{\text{ground}}$

Here,

$\Delta PE$ is the increase in potential energy

$P{E}_{\text{height}}$ is the potential energy at the height

$P{E}_{\text{ground}}$ is the potential energy at ground

Write the general expression for the potential energy.

$PE=mgh$

Here,

$m$ is the mass of the object

$g$ is the acceleration due to gravity

$h$ is the height

At ground, the potential energy is zero and thus, the expression for $\Delta PE$ becomes,

$\begin{array}{c}\Delta PE=mgh-0\\ =mgh\end{array}$

Substitute $5.0\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $1.8\text{m}$ for $h$ to find the increase in potential energy.

$\begin{array}{c}\Delta PE=\left(5.0\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(1.8\text{m}\right)\\ =88.2\text{J}\end{array}$

Conclusion:

Therefore, the increase in potential energy of the rock is $88.2\text{J}$.

(b)

To determine

The increase in kinetic energy to accelerate the rock horizontally from rest to the given speed.

Answer to Problem 11E

The increase in kinetic energy to accelerate the rock horizontally from rest to the given speed is $90.0\text{J}$.

Explanation of Solution

Given info: The mass of the rock is $5.0\text{kg}$ and the final velocity of the rock is $6.0\text{m/s}$.

Write the expression general expression for the kinetic energy.

$KE=\frac{1}{2}m{v}^2$

Here,

$KE$ is the kinetic energy

$m$ is the mass of the object

$v$ is the speed of the object

Write the expression for the increase in kinetic energy during acceleration of the object.

$\begin{array}{c}\Delta KE=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2\\ =\frac{1}{2}m\left(v_2^2-v_1^2\right)\end{array}$

Here,

$\Delta KE$ is the increase in kinetic energy

$v_1$ is the initial speed

$v_2$ is the final speed

Since the rock is accelerated from rest, the initial velocity is zero.

Substitute $5.0\text{kg}$ for $m$, $6.0\text{m/s}$ for $v_2$ and $0$ for $v_1$ to find the increase in kinetic energy $\Delta KE$.

$\begin{array}{c}\Delta KE=\frac{1}{2}\left(5.0\text{kg}\right)\left[{\left(6.0\text{m/s}\right)}^2-{\left(0\right)}^2\right]\\ =90.0\text{J}\end{array}$

Conclusion:

Therefore, the increase in kinetic energy to accelerate the rock horizontally from rest to the given speed is $90.0\text{J}$.

(c)

To determine

Which among the processes, lifting the rock $1.8\text{m}$ or accelerating it horizontally to a speed of $6.0\text{m/s}$ requires more work.

Answer to Problem 11E

Accelerating the rock horizontally to a speed of $6.0\text{m/s}$ requires more work.

Explanation of Solution

Given info: The increase in potential energy of the rock is $88.2\text{J}$ and the increase in kinetic energy to accelerate the rock horizontally from rest to the given speed is $90.0\text{J}$.

The work required to lift the rock from ground to a certain height is equal to the increase in potential energy of the rock at the height.

Therefore,

$\begin{array}{c}W=\Delta PE\\ =88.2\text{J}\end{array}$

The work required to accelerate the rock from rest to a finite velocity is equal to the increase in kinetic energy of the rock upon accelerating.

Therefore,

$\begin{array}{c}W=\Delta KE\\ =90.0\text{J}\end{array}$

Hence, accelerating the rock horizontally to a speed of $6.0\text{m/s}$ requires work of $90.0\text{J}$, which is higher than lifting it $1.8\text{m}$ that require $88.2\text{J}$ work.

Conclusion:

Thus, accelerating the rock horizontally to a speed of $6.0\text{m/s}$ requires more work.