Chapter 8.1, Problem 21PPSE

Solution

a.

To state: The equation that models the potential energy of a 2-kg skateboard that is sliding down a ramp. On Earth with a gravitational acceleration g , the potential energy stored in an object varies directly with its mass m and its vertical height h .

The resultant equation is $p=2gh$ .

Given information:

It is given that on Earth with a gravitational acceleration g , the potential energy stored in an object varies directly with its mass m and its vertical height h .

Explanation:

Let’s assume the Earth’s gravitational acceleration is g , the potential energy of object is p , mass is m and its vertical height h .

It is given that potential energy stored in an object varies directly with its mass m and its vertical height h . Then $p=gmh$ . Here g is constant.

Now substitute 2 for m in the equation $p=gmh$ .

  $\begin{array}{l}p=g\left(2\right)h\\ p=2gh\end{array}$

Therefore, the equation that models the potential energy of a 2-kg is $p=2gh$ .

b.

To state: The height of the ramp if skateboard has a potential energy of $-39.2{\text{kgm}}^2/{\text{s}}^2$ and the acceleration due to gravity is $g=-9.8{\text{m}}^2/{\text{s}}^2$ .

The resultant height is 2.

Given information:

From part (a), the equation that models the potential energy is $p=2gh$ . A potential energy of $-39.2{\text{kgm}}^2/{\text{s}}^2$ and the acceleration due to gravity is $g=-9.8{\text{m}}^2/{\text{s}}^2$ .

Explanation:

Consider the equation $p=2gh$ .

The given values of g and p are $-9.8\text{and}-39.2$ . Now substitute these values in equation $p=2gh$ to find the value of height.

  $\begin{array}{c}-39.2=2\left(-9.8\right)h\\ \frac{-39.2}{-19.6}=h\\ 2=h\end{array}$

Therefore, the height of the ramp is 2.