Unit 7: Energy

The same amount of work is required in both cases.

To lift an object at a constant speed, you must apply an upward lifting force equal to the weight of the object. For the two sacks in this question, the minimum forces required are \begin{equation} F_1 = 50\,\text{kg}\times g = 500\,\text{N} \end{equation} and \begin{equation} F_2 = 25\,\text{kg}\times g = 250\,\text{N} \end{equation} The work required to lift each sack is equal to the applied vertical force multiplied by the corresponding vertical displacement, which is calculated as follows: \begin{align} \text{Work on sack 1} &= 500\,\text{N}\times 2\,\text{m} \nonumber\\[6pt] &= 1000\,\text{J} \end{align} and \begin{align} \text{Work on sack 2} &= 250\,\text{N}\times 4\,\text{m} \nonumber\\[6pt] &= 1000\,\text{J} \end{align}

300 J

Based on the work–energy theorem, the net work done on the crate is equal to the change in its kinetic energy ($\Delta \text{KE}$). For the situation in this question (see Figure Q7.17), you have \begin{align} \Delta \text{KE} &= \text{net work} \nonumber\\[6pt] &= \text{net force} \times \text{distance} \nonumber\\[6pt] &= (100\,\text{N} - 70\,\text{N}) \times 10\,\text{m} \nonumber\\[6pt] &= 30\,\text{N} \times 10\,\text{m} \nonumber\\[6pt] &= 300\,\text{J} \end{align}

The initial potential energy transforms into kinetic energy during the fall, then into heat when the peach hits the ground.

While the peach is falling toward the ground, the force of gravity does work on the peach, increasing its kinetic energy. Almost all the initial gravitational potential energy converts into kinetic energy just before the peach hits the ground. A small amount of the total energy appears as heat due to air drag during the fall.

On impact, the majority of the peach’s kinetic energy transforms into thermal energy—the peach and the ground warm up a little. A small amount of the energy is carried by the sound waves that you hear upon impact.

A machine can multiply input force or input distance, but not input energy.

A specific amount of input energy can cause a machine to do a certain amount of work. The output force and distance, however, may vary, but their multiplication remains equal to the work done. For example, the machine can produce a small force that acts over a relatively long distance, or it can apply a large force over a shorter distance. It is important to note that a machine does not create energy. Therefore, output energy cannot exceed input energy.

150 N

The input work done at the end of the lever is equal to \begin{equation} F_\text{input} \times \text{distance} = 50\,\text{N} \times \text{distance} \end{equation} while the output work at the other end is given by \begin{equation} F_\text{output} \times \left(\frac{\text{distance}}{3}\right) \end{equation} Since energy is conserved, output work is equal to input work and you can calculate the maximum force exerted by the lever as follows: \begin{align} \text{Output work} &= \text{input work} \nonumber\\[6pt] \Rightarrow\quad F_\text{output} \times \left(\frac{\text{distance}}{3}\right) &= 50\,\text{N} \times \text{distance} \nonumber\\[6pt] \Rightarrow\quad F_\text{output} &= 3 \times 50\,\text{N} \nonumber\\[6pt] &= 150\,\text{N} \end{align}

100%

This perfect machine works at the maximum efficiency of \begin{equation} \frac{\text{Useful energy output}}{\text{Total energy input}} = 1 \end{equation}

The Sun.

Fossil fuels (coal and oil) were formed from remains of plants and other organisms on Earth hundreds of millions of years ago. As you know, visible light from the Sun is essential for photosynthesis, which causes plants to grow.

Rivers flow because of global water circulation powered by the Sun, which continuously vaporizes water from oceans and lakes and stores it in clouds. When rain falls on mountains, rivers flow and fill reservoirs behind dams.

Nonuniform heating of Earth’s surface by the Sun creates geographical differences in atmospheric pressure, which cause the winds to blow.

15 kJ

In this case, the movers must exert an upward force that is at least equal to the appliance’s weight: \begin{align} \text{Weight} &= (250\,\text{kg}) \, (10\,\text{m}/\text{s}^2) \nonumber\\[6pt] &= 2500\,\text{N} \end{align} Assuming a constant upward force, the work done on the appliance is calculated as \begin{align} \text{Work} &= (2500\,\text{N}) \, (6\,\text{m}) \nonumber\\[6pt] &= 15{,}000\,\text{J} \quad (\text{or}\; 15\,\text{kJ}) \end{align}

135 m

A car (of mass $m$) stops when its kinetic energy dissipates as heat due to the work done by the force of friction between the locked tires and the road. For this question, you can assume the force of friction ($F$) depends on the road and tire conditions only, independent of the car’s speed. Therefore, at both speeds mentioned in the question, the car experiences the same force while skidding.

By applying the work–energy theorem to the car traveling at an initial speed of 50 km/h, you can write

\begin{equation} F \times 15\,\text{m} = \frac{1}{2} m\, (50\,\text{km/h})^2 \label{ch07_q43_50} \end{equation}

The same car traveling at an initial speed of 150 km/h skids for a distance $d$ before stopping. Using the work–energy theorem, you have

\begin{align} F d &= \frac{1}{2} m\, (150\,\text{km/h})^2\nonumber\\[6pt] \Rightarrow\quad F d &= \frac{1}{2} m\, (3 \times 50\,\text{km/h})^2 \nonumber\\[6pt] \Rightarrow\quad F d &= 9 \times \frac{1}{2} m\, (50\,\text{km/h})^2 \label{ch07_q43_150} \end{align}

You can now substitute from Equation 7.20 into Equation 7.21 and calculate the distance $d$ as follows: \begin{align} F d &= 9 \times \left(F \times 15\,\text{m}\right) \nonumber\\[6pt] \Rightarrow\quad d &= 9 \times 15\,\text{m} \nonumber\\[6pt] \Rightarrow\quad d &= 135\,\text{m} \end{align} So when the speed triples, the stopping distance increases by nine times!

1000 N

This machine, sometimes called a hydraulic lever, transmits the pressure undiminished from one end to the other. According to the law of conservation of energy, the input work done on the small piston is equal to the output work performed by the large piston, which is expressed mathematically as \begin{align} (\text{force} \times \text{distance})_\text{input} &= (\text{force} \times \text{distance})_\text{output} \nonumber\\[6pt] \Rightarrow\quad 100\,\text{N} \times 10\,\text{cm} &= \text{weight} \times 1\,\text{cm} \end{align} Assuming 100% efficiency, you can calculate the maximum weight that the large piston can support as follows: \begin{align} \text{Weight} &= \frac{100\,\text{N} \times 10\,\text{cm}}{1\,\text{cm}} \nonumber\\[6pt] &= 10 \times 100\,\text{N} \nonumber\\[6pt] &= 1000\,\text{N} \end{align}

100 W

The work done on the crate is equal to \begin{equation} 50\,\text{N}\times 6.0\,\text{m} = 300\,\text{J} \end{equation} Since power is the work done (or energy spent) per second, you can write \begin{align} \text{Power} &= \frac{\text{work}}{\text{time}} \nonumber\\[6pt] &= \frac{300\,\text{J}}{3.0\,\text{s}} \nonumber\\[6pt] &= 100\,\text{W} \end{align}

Work is done inside the muscles of the person pushing against the wall.

Muscles consist of thousands of thin fibers that contract and relax quickly, even if this does not produce limb movement, such as when pushing against a stationary wall. No work is done on the wall, but a lot of work is done by the muscles inside the body, which causes them to eventually fatigue.

1. Kinetic energy is at its maximum at the instant the ball leaves the thrower’s hand.
2. Gravitational potential energy is at its maximum at the highest point of the ball’s ascent.

1. The ball starts at rest in the thrower’s hand. The thrower’s hand then exerts an upward force on the ball greater than the ball’s weight. As a result, the ball accelerates upward and reaches maximum speed (or greatest kinetic energy) as it leaves the thrower’s hand (see Figure Q7.80).
2. While in the air, the ball experiences free fall under the influence of gravity alone. During the upward stage of its motion, the ball slows down uniformly until it stops momentarily at the highest point. At this instant, the ball has zero kinetic energy and its gravitational potential energy is at its maximum value.

Yes, if you are moving as a single body.

For a single rigid object, zero momentum means zero speed and zero kinetic energy. However, a system of two or more bodies or objects can have zero net momentum while the different components of the system move and contribute to the system’s total kinetic energy. An example is a warm container filled with air that is sitting on a tabletop. In this case, the net momentum of the container is equal to zero, but the total kinetic energy of the air molecules inside it is relatively large.

No.

In this context, “to waste energy” means the machine transforms only a small fraction of energy input into useful forms of energy output. The remainder is wasted in forms of energy considered nonbeneficial. A machine, of course, cannot destroy energy.

The law of conservation of energy.

Ideally speaking, in the absence of friction and air resistance, a car would continue to move in the same direction and at a constant speed even if the engine were turned off (see Newton’s first law). In a real situation, when a car cruises at a constant speed, the engine works just to compensate for the energy lost due to air drag.

However, every time the driver stops the car at a traffic light, the initial kinetic energy of the car transforms entirely into heat energy through the work done by the braking system. When the light changes and the car pulls away, extra fuel must be burned to accelerate the car and regain the desired kinetic energy. Therefore, when cars must stop frequently at traffic lights, there is a significant increase in gas consumption, which is an important contributor to air pollution.

The ping-pong ball has a greater speed than the golf ball, and lighter molecules move faster than heavier ones in a gaseous mixture.

Recall the formula for kinetic energy: \begin{align} \text{KE} &= \frac{1}{2} m v^2 \nonumber\\[6pt] \Rightarrow\quad v &= \sqrt{\frac{2\, \text{KE}}{m}} \end{align} When two objects have the same kinetic energy, the heavier object (the golf ball in this case) moves slower than the lighter one (the ping-pong ball).

Similarly, you can argue that the lighter molecules in a gaseous mixture move faster than the heavier molecules, which have the same average kinetic energy.

The ball will return with less KE, and this does not contradict the law of energy conservation.

The decrease in the ball’s kinetic energy when the ball returns to its original vertical position is caused by the work done on the ball by the force of air resistance, which always acts opposite to the direction of motion. The air resistance converts part of the ball’s mechanical energy into heat, so the law of energy conservation is not violated.

Consider that gasoline was burned at an earlier time but not used to drive the car.

While idling or going downhill, a hybrid car burns gasoline to charge its electric battery, which is a part of a hybrid engine. The charged battery provides driving power (15 horsepower in this case) to assist the gasoline engine when required. The benefit of a hybrid engine is that it is more efficient than a regular gasoline engine.

Think about the difference between the total energy produced by the electric company in one day and the maximum energy that can be delivered to customers in one hour.

The problem here is that the electric company can’t produce energy at a sufficient rate during a period of high demand. Since “power” refers to the amount of energy produced or delivered per unit time, “power crisis” is a more accurate description of the situation in this question.

The block that is twice as massive has twice as much potential energy as the other block.

When raised to an elevation $h$ above the ground, an object of mass $m$ gains gravitational potential energy as given by the equation \begin{equation} \text{PE} = mgh \end{equation} The potential energy is proportional to the mass of the raised object. So if the elevation is kept constant, doubling the mass also doubles the potential energy. Another way to look at this problem is to think of the heavier block as two lighter blocks raised together to the same elevation.

The kinetic energy and potential energy are converted primarily to heat.

Work done by air resistance transforms the mechanical (kinetic and potential) energy of the airplane into heat, which results in an increase of the temperature of the plane’s body and surrounding air. (Note that a shooting star is a piece of rock that burns because of the intense heat generated when it rubs air molecules in Earth’s atmosphere.) A small fraction of the initial mechanical energy is converted to the sound waves you hear as the plane lands.

No work is done on the satellite, whose speed remains constant.

As the satellite moves along a circular orbit (see Figure E7.C), the force of gravity ($F_g$) continuously acts perpendicular to the satellite’s velocity ($v$). In this case, $F_g$ is actually the centripetal force responsible for the uniform circular motion. Since the satellite does not perform any displacement in the direction of the applied force, no work is done by the force of gravity on the satellite, whose kinetic energy (and speed) does not change with time.

Momentum is conserved, but kinetic energy is not conserved.

Initially, the two cars move with equal and opposite momenta, so the total momentum of the system before the collision is equal to zero. After the collision, the vehicles stick together and come to rest (zero final momentum).

The total kinetic energy of the two-car system is equal to the sum of the kinetic energies of the moving cars. After the accident, the cars stop moving and their individual kinetic energies become zero. So, during impact, the initial kinetic energies transform into heat (i.e., kinetic energy of molecular motion). Note, however, that the total energy of the system is conserved.

Open windows on a car traveling at relatively high speeds cause a significant increase in air drag.

Opening the windows of a moving car causes turbulent movement of the air inside the car and the surrounding air. This reduces the streamlined movement of the air around the vehicle, significantly increases air drag, especially at higher speeds. So, to maintain your speed, you must step on the gas more, which causes the car to consume more fuel than it would if the windows were closed.