Chapter 6, Problem 5SP

(a)

To determine

Whether the sled will make it to the top of the second hump if no kinetic energy is given to the sled at the start of its motion.

Answer to Problem 5SP

Even if no kinetic energy is given to the sled at the start of its motion, the sled will make it up to the top of the second hump.

Explanation of Solution

Given info: The height of the first point is $40\text{m}$ and that of second point is $30\text{m}$, the combined mass of the sled and the rider is $40\text{kg}$ and the work done against friction is $2000\text{J}$.

Write the expression for the gravitational potential energy.

$PE=mgh$ (1)

Here,

$PE$ is the gravitational potential energy

$m$ is the mass

$g$ is the acceleration due to gravity

$h$ is the height

The value of $g$ is $9.8{\text{m/s}}^{\text{2}}$.

Substitute $40\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $40\text{m}$ for $h$ in equation (1) to find the potential energy at the first point.

$\begin{array}{c}P{E}_1=\left(40\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(40\text{m}\right)\\ =15680\text{J}\end{array}$

Here,

$P{E}_1$ is the potential energy at the first point

Substitute $40\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $30\text{m}$ for $h$ in equation (1) to find the potential energy at second point.

$\begin{array}{c}P{E}_2=\left(40\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(30\text{m}\right)\\ =11760\text{J}\end{array}$

Here,

$P{E}_2$ is the potential energy at the second hump

Find the difference in potential energy of the two points.

$\begin{array}{c}P{E}_1-P{E}_2=15680\text{J}-11760\text{J}\\ =3,920\text{ J}\end{array}$

The difference in potential energy of the two points is $3920\text{J}$, while the loss of energy due to friction is only $2000\text{J}$. Thus, even if no kinetic energy is given to the sled at the start of its motion, the sled will make it up to the top of the second hump.

Conclusion:

Therefore even if no kinetic energy is given to the sled at the start of its motion, the sled will make it up to the top of the second hump.

(b)

To determine

The maximum height that the second hump could be in order for the sled to reach the top.

Answer to Problem 5SP

In order for the sled to reach the top of the second hump, the maximum height it can have is $34.9\text{m}$.

Explanation of Solution

Given info: The work done against friction is $2000\text{J}$.

While moving, the potential energy of the system is converted to kinetic energy. The maximum height the second hump can have corresponds to the maximum kinetic energy gained during the motion. The maximum kinetic energy gained will be equal to the potential energy at the second hump.

$\begin{array}{c}KE=P{E}_2\\ P{E}_2=P{E}_1-W_{\text{f}}\end{array}$ (2)

Here,

$KE$ is the kinetic energy

$W_{\text{f}}$ is the work done against friction

Assume $h^{\prime }$ is the maximum height the second hump can have.

$P{E}_2=mg{h}^{\prime }$ (3)

Put equation (3) in (2) and solve for $h^{\prime }$.

$\begin{array}{c}mg{h}^{\prime }=P{E}_1-W_{\text{f}}\\ h^{\prime }=\frac{P{E}_1-W_{\text{f}}}{mg}\end{array}$

Substitute $15680\text{J}$ for $P{E}_1$, $2000\text{J}$ for $W_{\text{f}}$, $40\text{kg}$ for $m$ and $9.8{\text{m/s}}^{\text{2}}$ for $g$ in the above equation to find $h^{\prime }$.

$\begin{array}{c}{h}^{\prime }=\frac{15680\text{J}-2000\text{J}}{\left(40\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)}\\ =34.9\text{m}\end{array}$

Conclusion:

Therefore in order for the sled to reach the top of the second hump, the maximum height it can have is $34.9\text{m}$.