Question

An object whose mass is \(0.5 \mathrm{~kg}\) has a velocity of \(30 \mathrm{~m} / \mathrm{s}\). Determine (a) the final velocity, in \(\mathrm{m} / \mathrm{s}\), if the kinetic energy of the object decreases by \(130 \mathrm{~J}\). (b) the change in elevation, in \(\mathrm{ft}\), associated with a \(130 \mathrm{~J}\) change in potential energy. Let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

Short answer

(a) Final velocity is 19.49 m/s. (b) Change in elevation is 86.89 ft.

Step-by-step solution

Determine Initial Kinetic Energy

First, use the formula for kinetic energy: \[ KE = \frac{1}{2}mv^2 \] where \(m\) is mass and \(v\) is velocity. Here, \(m = 0.5\, \mathrm{kg}\) and \(v = 30\, \mathrm{m/s}\). Substitute the values: \[ KE = \frac{1}{2}(0.5)(30^2) = 0.25(900) = 225\, \mathrm{J} \]

Calculate Final Kinetic Energy

Since the kinetic energy decreases by 130\,J, subtract this value from the initial kinetic energy: \[ KE_{\text{final}} = 225\, \mathrm{J} - 130\, \mathrm{J} = 95\, \mathrm{J} \]

Determine Final Velocity

Use the kinetic energy formula in reverse to solve for velocity: \[ KE = \frac{1}{2}mv^2 \] Rearrange to solve for the final velocity \(v_{\text{final}}\): \[ v_{\text{final}} = \sqrt{\frac{2(KE_{\text{final}})}{m}} \] Substitute \(KE_{\text{final}} = 95\, \mathrm{J}\) and \(m = 0.5\, \mathrm{kg}\): \[ v_{\text{final}} = \sqrt{\frac{2(95)}{0.5}} = \sqrt{380} = 19.49\, \mathrm{m/s} \]

Determine Change in Elevation

Use the potential energy formula: \[ PE = mgh \] Rearrange to solve for the change in height \(\Delta h\): \[ \Delta h = \frac{\Delta PE}{mg} \] Given \(\Delta PE = 130\, \mathrm{J}\), \(m = 0.5\, \mathrm{kg}\), and \(g = 9.81\, \mathrm{m/s^2}\), substitute the values: \[ \Delta h = \frac{130}{0.5 \times 9.81} = \frac{130}{4.905} = 26.5 \, \mathrm{m} \] Convert \(\Delta h\) from meters to feet (1 meter = 3.281 feet): \[ \Delta h = 26.5 \, \mathrm{m} \times 3.281 \frac{\mathrm{ft}}{\mathrm{m}} = 86.89 \, \mathrm{ft} \]

Key concepts

Kinetic Energy Calculation

Kinetic energy is the energy possessed by an object due to its motion. It's given by the formula: \[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass and \( v \) is the velocity of the object. For the initial calculation:

• Given mass \( m = 0.5 \, \mathrm{kg} \)
• Given velocity \( v = 30 \, \mathrm{m/s} \)

Using the formula, we calculate the initial kinetic energy as:\[ KE = \frac{1}{2}(0.5)(30^2) = 0.25(900) = 225 \, \mathrm{J} \]
This initial kinetic energy plays a crucial role in understanding the energy changes and conversions in subsequent steps.

Potential Energy Change

Potential energy is the energy possessed by an object due to its position or height. The formula for potential energy is:\[ PE = mgh \]where \( m \) is mass, \( g \) is the gravitational acceleration, and \( h \) is the height. To find the change in elevation corresponding to a \( 130 \ \mathrm{J} \) change in potential energy:

• Given \( \Delta PE = 130 \ \mathrm{J} \)
• Mass \( m = 0.5 \ \mathrm{kg} \)
• Gravitational acceleration \( g = 9.81 \ \mathrm{m/s^2} \)

We rearrange the formula to find the change in height \( \Delta h \):\[ \Delta h = \frac{\Delta PE}{mg} \]Substituting the values:\[ \Delta h = \frac{130}{0.5 \times 9.81} = \frac{130}{4.905} = 26.5 \, \mathrm{m} \]To convert this height from meters to feet:\[ \Delta h = 26.5 \, \mathrm{m} \times 3.281 \frac{\mathrm{ft}}{\mathrm{m}} = 86.89 \ \mathrm{ft} \]

Energy Conversion

Energy conversion refers to the process of changing energy from one form to another. This exercise demonstrates the conversion between kinetic and potential energy.
Initially, we had a kinetic energy of 225 \( \mathrm{J} \). With a decrease of 130 \( \mathrm{J} \), the new kinetic energy becomes:\[ KE_{\text{final}} = 225 \ \mathrm{J} - 130 \ \mathrm{J} = 95 \ \mathrm{J} \]
This reduction in kinetic energy directly affects the velocity of the object.
Furthermore, the problem shows how potential energy change is linked to the height or elevation, allowing the conversion formula to be applied, demonstrating a practical scenario of energy transformation.
Understanding how energy changes forms helps in analyzing real-world movements and mechanisms.

Physics of Motion

Physics of motion involves analyzing how objects move under different forces and conditions. Core concepts include kinetic and potential energy, velocity, and acceleration.
Using the kinetic energy formula:\[ KE = \frac{1}{2} mv^2 \]and solving for final velocity:\
\[ v_{\text{final}} = \sqrt{\frac{2 KE_{\text{final}}}{m}} \]Given \( KE_{\text{final}} = 95 \ \mathrm{J} \) and \( m = 0.5 \ \mathrm{kg} \):\[ v_{\text{final}} = \sqrt{\frac{2(95)}{0.5}} = \sqrt{380} = 19.49 \ \mathrm{m/s} \]
This shows the close relationship between energy metrics and their direct effects on motion parameters such as velocity. Analyzing these relationships allows us to predict and calculate the outcomes of various physical scenarios.