Chapter 7, Problem 12P

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m. (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate–incline system owing to friction. (c) How much work is done by the 100-N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.00 m?

To determine

To determine: The work done by the gravitational force on the crate.

Answer to Problem 12P

Answer: The work done by the gravitational force on the crate is $-\text{168}\text{J}$.

Explanation of Solution

Explanation:

Given information:

A crate of mass $10\text{kg}$ is pulled up a rough incline with an initial speed of $1.5\text{m}/\text{s}$ and the pulling force is $100\text{N}$ parallel to the incline, which makes an angle of $20.0°$ with the horizontal. The coefficient of kinetic friction is $0.4$, and the crate is pulled $5.00\text{m}$.

The free body diagram of the system is shown below.

Figure I

The vertical distance moved by the crate is,

$h=d\sin \theta$

• $h$ is the vertical distance.
• $d$ is the distance moved by crate along the incline.

The formula to calculate the work done by gravitational force is,

$W_{\text{g}}=mgh$

Substitute $d\sin \theta$ for $h$ in the above equation.

$W_{\text{g}}=mgd\sin \theta$

Substitute $10\text{kg}$ for $m$, $-9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $20°$ for $\theta$ and $5.00\text{m}$ for $d$ to find $W_{\text{g}}$.

$\begin{array}{c}{W}_{\text{g}}=10\text{kg}\times \left(-9.81\text{m}/{\text{s}}^{\text{2}}\right)\times 5.00\text{m}\times \sin \left(20°\right)\\ \approx -\text{168}\text{J}\end{array}$

Conclusion:

Therefore, the work done by the gravitational force on the crate is $-\text{168}\text{J}$.

To determine

To determine: The increase in internal energy of the crate-incline system owing to friction.

Answer to Problem 12P

Answer: The increase in internal energy of the crate-incline system owing to friction is $\text{184}\text{J}$.

Explanation of Solution

Explanation:

Given information:

A crate of mass $10\text{kg}$ is pulled up a rough incline with an initial speed of $1.5\text{m}/\text{s}$ and the pulling force is $100\text{N}$ parallel to the incline, which makes an angle of $20.0°$ with the horizontal. The coefficient of kinetic friction is $0.4$, and the crate is pulled $5.00\text{m}$.

Formula to calculate kinetic frictional force is,

$f_{\text{k}}=\mu N$

• $\mu$ is the kinetic frictional coefficient.
• $N$ is the normal force between crate and plane.
• $f_{\text{k}}$ is the kinetic frictional force.

Formula to calculate the increase in the internal energy is,

$E=f_{\text{k}}d$

• $E$ is the internal energy of the crate.

Substitute $\mu N$ for $f_{\text{k}}$ and $mg\cos \theta$ for $N$ in above equation.

$E=\mu mg\cos \theta \times d$

Substitute $10\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $20°$ for $\theta$, $0.4$ for $\mu$ and $5.00\text{m}$ for $d$ to find $E$.

$\begin{array}{c}E=0.4\times 10\text{kg}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times \cos 20°\times 5.00\text{m}\\ =\text{184}\text{J}\end{array}$

Conclusion:

Therefore, the increase in internal energy of the crate-incline system owing to friction is $\text{184}\text{J}$.

To determine

To determine: The work done by the $100\text{N}$ force on the crate.

Answer to Problem 12P

Answer: The work done by the $100\text{N}$ force on the crate is $\text{500}\text{J}$.

Explanation of Solution

Explanation:

Given information:

A crate of mass $10\text{kg}$ is pulled up a rough incline with an initial speed of $1.5\text{m}/\text{s}$ and the pulling force is $100\text{N}$ parallel to the incline, which makes an angle of $20.0°$ with the horizontal. The coefficient of kinetic friction is $0.4$, and the crate is pulled $5.00\text{m}$.

Formula to calculate work done on the crate is,

$W=F\times d$

• $F$ is the force applied on the crate.

Substitute $100\text{N}$ for $F$ and $5.00\text{m}$ for $d$ to find $W$.

$\begin{array}{c}W=100\text{N}\times 5.00\text{m}\\ =\text{500}\text{J}\end{array}$

Conclusion:

Therefore, the work done by the $100\text{N}$ force on the crate is $\text{500}\text{J}$.

To determine

To determine: The change in kinetic energy of the crate.

Answer to Problem 12P

Answer: The change in kinetic energy of the crate is $148\text{J}$.

Explanation of Solution

Explanation:

Given information:

A crate of mass $10\text{kg}$ is pulled up a rough incline with an initial speed of $1.5\text{m}/\text{s}$ and the pulling force is $100\text{N}$ parallel to the incline, which makes an angle of $20.0°$ with the horizontal. The coefficient of kinetic friction is $0.4$, and the crate is pulled $5.00\text{m}$.

Formula to calculate change in kinetic energy of the crate is,

$\Delta KE=W_{\text{g}}+W-E$

• $\Delta KE$ is the change in kinetic energy of the crate.

Substitute $-168\text{J}$ for $W_{\text{g}}$, $500\text{J}$ for $W$ and $184\text{J}$ for $E$ to find $\Delta KE$.

$\begin{array}{c}\Delta KE=-168\text{J}+500\text{J}-184\text{J}\\ =148\text{J}\end{array}$

Conclusion:

Therefore, the change in kinetic energy of the crate is $148\text{J}$.

To determine

To determine: The speed of the crate after being pulled $5.00\text{m}$.

Answer to Problem 12P

Answer: The speed of the crate after being pulled $5.00\text{m}$ is $5.64\text{m}/\text{s}$.

Explanation of Solution

Explanation:

Given information:

A crate of mass $10\text{kg}$ is pulled up a rough incline with an initial speed of $1.5\text{m}/\text{s}$ and the pulling force is $100\text{N}$ parallel to the incline, which makes an angle of $20.0°$ with the horizontal. The coefficient of kinetic friction is $0.4$, and the crate is pulled $5.00\text{m}$.

Formula to calculate change in kinetic energy of the crate is,

$\Delta KE=\frac{1}{2}m{v}_{\text{f}}{}^2-\frac{1}{2}m{v}_{\text{i}}{}^2$

• $v_{\text{f}}$ is the final speed of the crate.
• $v_{\text{i}}$ is the initial speed of the crate.

Substitute $148\text{J}$ for $\Delta KE$, $10\text{kg}$ for $m$ and $1.5\text{m}/\text{s}$ for $v_{\text{i}}$ to find $v_{\text{f}}$.

$\begin{array}{c}148\text{J}=\frac{1}{2}\times 10\text{kg}\times v_{\text{f}}{}^2-\frac{1}{2}\times 10\text{kg}\times {\left(1.5\text{m}/\text{s}\right)}^2\\ v_{\text{f}}=\sqrt{\frac{2\times 148\text{J}}{10\text{kg}}+{\left(1.5\text{m}/\text{s}\right)}^2}\\ =5.64\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the crate after being pulled $5.00\text{m}$ is $5.64\text{m}/\text{s}$.