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Chapter 2: Problem 3

An object whose weight is 100 lbf experiences a decrease in kinetic energy of \(500 \mathrm{ft}\) - lbf and an increase in potential energy of \(1500 \mathrm{ft}\) - lbf. The initial velocity and elevation of the object, each relative to the surface of the earth, are \(40 \mathrm{ft} / \mathrm{s}\) and \(30 \mathrm{ft}\), respectively. If \(g=32.2 \mathrm{ft} / \mathrm{s}^{2}\), determine (a) the final velocity, in \(\mathrm{ft} / \mathrm{s}\). (b) the final elevation, in \(\mathrm{ft}\).

Short Answer

Expert verified
The final velocity is approximately 35.8 ft/s and the final elevation is approximately 45.02 ft.

Step by step solution

01

Understand the given data

The object's initial weight is 100 lbf. It experiences a decrease in kinetic energy of 500 ft-lbf and an increase in potential energy of 1500 ft-lbf. Its initial velocity is 40 ft/s and initial elevation is 30 ft. Acceleration due to gravity is given as 32.2 ft/s^2.
02

Calculate the initial kinetic energy

Use the formula for kinetic energy: \[ KE = \frac{1}{2}mv^2 \] First, find the mass, which can be derived from the weight: \[ W = mg \rightarrow m = \frac{W}{g} \] Substitute the given values: \[ m = \frac{100}{32.2} \approx 3.11 \text{ slugs} \] Now calculate the initial kinetic energy: \[ KE_1 = \frac{1}{2} \times 3.11 \times 40^2 = 2488 \text{ ft-lbf} \]
03

Calculate the final kinetic energy

The kinetic energy decreases by 500 ft-lbf: \[ KE_2 = KE_1 - 500 \] \[ KE_2 = 2488 - 500 = 1988 \text{ ft-lbf} \]
04

Determine the final velocity

Use the final kinetic energy to find the final velocity: \[ KE_2 = \frac{1}{2}mv^2 \rightarrow v = \sqrt{\frac{2KE_2}{m}} \] Substitute the values: \[ v = \sqrt{\frac{2 \times 1988}{3.11}} \approx 35.8 \text{ ft/s} \]
05

Calculate the initial potential energy

Use the formula for potential energy: \[ PE = mgh \] Substitute the initial values: \[ PE_1 = 3.11 \times 32.2 \times 30 = 3000.6 \text{ ft-lbf} \]
06

Calculate the final potential energy

The potential energy increases by 1500 ft-lbf: \[ PE_2 = PE_1 + 1500 \] \[ PE_2 = 3000.6 + 1500 = 4500.6 \text{ ft-lbf} \]
07

Determine the final elevation

Use the final potential energy to find the final elevation: \[ PE_2 = mgh \rightarrow h = \frac{PE_2}{mg} \] Substitute the values: \[ h = \frac{4500.6}{3.11 \times 32.2} \approx 45.02 \text{ ft} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It can be calculated using the formula: \( KE = \frac{1}{2}mv^2 \). Here, \( m \) denotes the mass and \( v \) is the velocity of the object. In the exercise, we calculated the initial kinetic energy by first finding the object's mass using its weight and the acceleration due to gravity. The initial kinetic energy came out as 2488 ft-lbf based on the given initial velocity of 40 ft/s.
Potential Energy
Potential energy is the energy possessed by an object due to its position or height. It's given by the formula: \[ PE = mgh \]. Here, \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) represents the height. In this problem, we calculated the initial potential energy of the object at a height of 30 ft. Subsequent calculations showed an increase of 1500 ft-lbf in potential energy, helping us determine the final potential energy.
Final Velocity
The final velocity of an object is its speed at the end of the motion under consideration. This can be derived from the final kinetic energy: \[ KE = \frac{1}{2}mv^2 \rightarrow v = \sqrt{\frac{2KE}{m}} \]. For the object in question, the final kinetic energy was calculated to be 1988 ft-lbf, leading to a final velocity of approximately 35.8 ft/s. This is less than the initial velocity, reflecting the decrease in kinetic energy.
Final Elevation
The final elevation indicates the height of the object after experiencing a change in energy. It can be calculated from the final potential energy: \[ PE = mgh \rightarrow h = \frac{PE}{mg} \]. Using the exercise data, we found the final potential energy to be 4500.6 ft-lbf. Applying this to our formula, the final elevation was determined to be about 45.02 ft, higher than the initial elevation of 30 ft due to the increase in potential energy.
Energy Conservation
The principle of energy conservation states that the total energy in an isolated system remains constant. Energy can neither be created nor destroyed, but it can change forms between kinetic and potential energy. In the given problem, the object鈥檚 decrease in kinetic energy coincided with an increase in potential energy, demonstrating this principle. Even though the forms of energy changed, the total energy balance remained consistent, reinforcing the law of conservation of energy.

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Most popular questions from this chapter

For a refrigerator with automatic defrost and a topmounted freezer, the annual cost of electricity is \(\$ 55\). (a) Evaluating electricity at 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\), determine the refrigerator's annual electricity requirement, in \(\mathrm{kW} \cdot \mathrm{h}\). (b) If the refrigerator's coefficient of performance is 3 , determine the amount of energy removed from its refrigerated space annually, in MJ.

A closed system of mass \(10 \mathrm{~kg}\) undergoes a process during which there is energy transfer by work from the system of \(0.147 \mathrm{~kJ}\) per \(\mathrm{kg}\), an elevation decrease of \(50 \mathrm{~m}\), and an increase in velocity from \(15 \mathrm{~m} / \mathrm{s}\) to \(30 \mathrm{~m} / \mathrm{s}\). The specific internal energy decreases by \(5 \mathrm{~kJ} / \mathrm{kg}\) and the acceleration of gravity is constant at \(9.7 \mathrm{~m} / \mathrm{s}^{2}\). Determine the heat transfer for the process, in \(\mathrm{kJ}\).

The two major forces opposing the motion of a vehicle moving on a level road are the rolling resistance of the tires, \(F_{\mathrm{r}}\), and the aerodynamic drag force of the air flowing around the vehicle, \(F_{\mathrm{d}}\), given respectively by $$ F_{\mathrm{r}}=f \mathrm{~W}, \quad F_{\mathrm{d}}=C_{\mathrm{d}} \mathrm{A} \frac{1}{2} \rho \mathrm{V}^{2} $$ where \(f\) and \(C_{\mathrm{d}}\) are constants known as the rolling resistance coefficient and drag coefficient, respectively, \(W\) and \(A\) are the vehicle weight and projected frontal area, respectively, V is the vehicle velocity, and \(\rho\) is the air density. For a passenger car with \(W=3040\) lbf, \(A=6.24 \mathrm{ft}^{2}\), and \(C_{\mathrm{d}}=0.25\), and when \(f=0.02\) and \(\rho=0.08 \mathrm{lb} / \mathrm{ft}^{3}\) (a) determine the power required, in hp, to overcome rolling resistance and aerodynamic drag when \(\mathrm{V}\) is \(55 \mathrm{mi} / \mathrm{h}\). (b) plot versus vehicle velocity ranging from 0 to \(75 \mathrm{mi} / \mathrm{h}\) (i) the power to overcome rolling resistance, (ii) the power to overcome aerodynamic drag, and (iii) the total power, all in hp. What implication for vehicle fuel economy can be deduced from the results of part (b)?

Vehicle crumple zones are designed to absorb energy during an impact by deforming to reduce transfer of energy to occupants. How much kinetic energy, in Btu, must a crumple zone absorb to fully protect occupants in a 3000 -lb vehicle that suddenly decelerates from \(10 \mathrm{mph}\) to \(0 \mathrm{mph}\) ?

The drag force, \(F_{\mathrm{d}}\), imposed by the surrounding air on a vehicle moving with velocity \(\mathrm{V}\) is given by $$ F_{\mathrm{d}}=C_{\mathrm{d}} \mathrm{A}_{2} \rho \mathrm{V}^{2} $$ where \(C_{\mathrm{d}}\) is a constant called the drag coefficient, A is the projected frontal area of the vehicle, and \(\rho\) is the air density. Determine the power, in hp, required to overcome aerodynamic drag for an automobile moving at (a) 25 miles per hour, (b) 70 miles per hour. Assume \(C_{\mathrm{d}}=0.28, \mathrm{~A}=25 \mathrm{ft}^{2}\), and \(\rho=0.075 \mathrm{lb} / \mathrm{ft}^{2}\).
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