Question

potential energy challenge
objective: students will be able to demonstrate their knowledge of potential energy in order to collaboratively utilize lab equipment to measure and calculate the energy of various objects.

potential energy is the stored energy of position possessed by an object

an object can store energy as the result of its position. for example, the heavy ball of a demolition machine is storing energy when it is held at an elevated position. this stored energy of position is referred to as potential energy. similarly, a drawn bow is able to store energy as the result of its position. when assuming its usual position (i.e., when not drawn), there is no energy stored in the bow. yet when its position is altered from its usual equilibrium position, the bow is able to store energy by virtue of its position. this stored energy of position is referred to as potential energy.

potential energy:

1. definition: potential energy is energy due to the ________ of the object. the object is not in ________.
2. potential energy formula: pe =
3. m = ______ g = ______ and h = ________
4. the unit for pe is ____

practical problem:
a 3458 g ball is on a 60 cm coffee table. calculate the potential energy of the ball.

m = ______ g = ______ and h ____
a. pe = (write the formula)
b. pe = (substitute the values)
c. pe = (write final answer, include unit)

recall:
two tools to measure mass are:

a.
b.

Explanation:

Step1: Identify the formula for potential energy

The formula for gravitational potential energy is \( PE = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)), and \( h \) is height.

Step2: Convert units

• Mass \( m = 3458 \, \text{g} = \frac{3458}{1000} = 3.458 \, \text{kg} \) (since \( 1 \, \text{kg} = 1000 \, \text{g} \))
• Height \( h = 60 \, \text{cm} = \frac{60}{100} = 0.6 \, \text{m} \) (since \( 1 \, \text{m} = 100 \, \text{cm} \))
• \( g = 9.8 \, \text{m/s}^2 \) (standard value for acceleration due to gravity near Earth's surface)

Step3: Substitute values into the formula

\( PE = mgh = 3.458 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.6 \, \text{m} \)

Step4: Calculate the result

First, multiply \( 3.458 \times 9.8 = 33.8884 \)
Then, multiply \( 33.8884 \times 0.6 = 20.33304 \)
So, \( PE \approx 20.33 \, \text{J} \) (Joules, since the unit of potential energy is Joule, \( 1 \, \text{J} = 1 \, \text{kg·m}^2/\text{s}^2 \))

Answer:

For the Potential Energy section:

1. Definition: Potential energy is energy due to the \(\boldsymbol{\text{position}}\) of the object. The object is not in \(\boldsymbol{\text{motion}}\).
2. Potential Energy formula: \( PE = \boldsymbol{mgh} \)
3. \( m = \boldsymbol{\text{mass}} \), \( g = \boldsymbol{\text{acceleration due to gravity (9.8 m/s}^2\text{)}} \), and \( h = \boldsymbol{\text{height}} \)
4. The unit for PE is \(\boldsymbol{\text{Joule (J)}}\)

For the Practice Problem:

\( m = \boldsymbol{3.458 \, \text{kg}} \), \( g = \boldsymbol{9.8 \, \text{m/s}^2} \), and \( h = \boldsymbol{0.6 \, \text{m}} \)
a. \( PE = \boldsymbol{mgh} \)
b. \( PE = 3.458 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.6 \, \text{m} \)
c. \( PE = \boldsymbol{20.33 \, \text{J}} \) (or more precisely \( 20.33304 \, \text{J} \))