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Chapter 8: Problem 28

An electric scooter has a battery capable of supplying 120 \(\mathrm{Wh}\) of energy. If friction forces and other losses account for 60.0\(\%\) of the energy usage, what altitude change can a rider achieve when driving in hilly terrain, if the rider and scooter have a combined weight of 890 \(\mathrm{N} ?\)

Short Answer

Expert verified
The altitude change is approximately \(17.76 \text{ meters}.\)

Step by step solution

01

Determine the Usable Energy

The total energy supplied by the battery is 120 \(\mathrm{Wh}\), but only 40\% (100\% - 60\%) of it is used for altitude change due to energy losses from friction forces and other causes. To find the usable energy in joules, multiply the energy in Wh by the conversion factor of 3600 joules per Wh and by the efficiency factor (0.40), resulting in \(E_{usable} = 120 \mathrm{Wh} \times 3600 \frac{\text{joules}}{\text{Wh}} \times 0.40\).
02

Calculate Altitude Change

The altitude change can be calculated using the work-energy principle. The work done against gravity equals the change in gravitational potential energy which is \(mg\Delta h = E_{usable}\), where \(m\) is the mass of the rider and scooter system, \(g\) is acceleration due to gravity (approximately \(9.81 \frac{\text{m}}{\text{s}^2}\)), and \(\Delta h\) is the altitude change. First, calculate the mass \(m\) by dividing the weight (890 \(\mathrm{N}\)) by \(g\). Then, solve for \(\Delta h\) using the formula \(\Delta h = \frac{E_{usable}}{mg}\).
03

Compute the Actual Value

Substitute the values into the equations from the previous steps and solve for \(\Delta h\). First, calculate \(E_{usable} = 120 \times 3600 \times 0.40 = 172800 \text{joules}\). Then determine the mass: \(m = \frac{890 \mathrm{N}}{9.81 \frac{\text{m}}{\text{s}^2}}\). Finally use the calculated mass to find \(\Delta h = \frac{172800 \text{joules}}{890 \mathrm{N}}\). Remember that since \(\mathrm{N} = \mathrm{kg} \cdot \frac{\text{m}}{\text{s}^2}\), the units will simplify appropriately to give the altitude change in meters.

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Key Concepts

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

Gravitational Potential Energy
Gravitational Potential Energy (GPE) is the energy an object possesses due to its position in a gravitational field. The higher the object is above a reference point, the more gravitational potential energy it has. This energy is given by the formula:
\[ GPE = mgh \]
where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity, and \(h\) is the height above the reference point. In the context of the problem with an electric scooter, when considering an altitude change, we're essentially measuring the change in the scooter's gravitational potential energy. The scooter gains GPE when it ascends hills and loses GPE when it descends, according to the work done against gravity.
Helpful advice for students is to make sure they understand that weight (\(W\text{, in Newtons}\)) can be used to find mass by dividing by the acceleration due to gravity (\(W = mg\)), because mass is a component in calculating gravitational potential energy.
Energy Efficiency
Energy efficiency is a measure of how much of the total energy input to a system is used for its intended task versus how much is lost in the process. For example, in the case of the electric scooter, not all energy from the battery is used for altering altitude 鈥� some is lost to friction and other factors. The loss of energy to these factors is often represented as a percentage. In the problem given, 60% of the energy is lost, meaning the efficiency is 40%. To calculate the usable energy (\(E_{usable}\)), you multiply the total energy by the efficiency (as a decimal):
\[ E_{usable} = E_{total} \times \text{Efficiency} \]
It鈥檚 important to underline the concept to students that efficiency will always be less than 100%, since no practical system can convert energy without any losses.

Applying Energy Efficiency to Solve Problems


In solving problems, students must remember to apply the efficiency factor to determine the actual usable energy available for the task. This step often involves converting from percentages to a decimal factor.
Energy Conversion
Energy conversion involves transforming one form of energy into another. This is a fundamental concept in physics, as it underpins many processes in daily life. For instance, in the electric scooter scenario, the chemical energy stored in the battery is converted into mechanical energy, part of which is used to ascend to a higher altitude (increasing gravitational potential energy).
However, during this energy conversion process, not all energy is perfectly converted to the desired form - some of it is lost, mainly due to friction and other resistances. This is highlighted in the problem where only 40% of the battery's energy is usable for gaining altitude.

Understanding Practical Energy Conversion


By grasping the concept of energy conversion, students can better understand the significance of energy losses (such as friction) and the importance of efficiency. It鈥檚 also crucial to note that the law of conservation of energy states energy cannot be created or destroyed, only changed from one form to another - a fact students should always bear in mind when thinking about energy conversions.

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Most popular questions from this chapter

At \(11 : 00\) A.M. on September \(7,2001\) , more than 1 million British school children jumped up and down for one minute. The curriculum focus of the "Giant Jump" was on earthquakes, but it was integrated with many other topics, such as exercise, geography, cooperation, testing hypotheses, and setting world records. Children built their own seismographs, which registered local effects. (a) Find the mechanical energy released in the experiment. Assume that 1050000 children of average mass 36.0 kg jump twelve times each, raising their centers of mass by 25.0 \(\mathrm{cm}\) each time and briefly resting between one jump and the next. The free-fall acceleration in Britain is 9.81 \(\mathrm{m} / \mathrm{s}^{2}\) . (b) Most of the energy is converted very rapidly into internal energy within the bodies of the children and the floors of the school buildings. Of the energy that propagates into the ground, most produces high-frequency "microtremor" vibrations that are rapidly damped and cannot travel far. Assume that 0.01\(\%\) of the energy is carried away by a long-range seismic wave. The magnitude of an earthquake on the Richter scale is given by $$M=\frac{\log E-4.8}{1.5}$$ where \(E\) is the seismic wave energy in joules. According to this model, what is the magnitude of the demonstration quake? (It did not register above background noise over-seas or on the seismograph of the Wolverton Seismic Vault, Hampshire.)

A boy in a wheelchair (total mass 47.0 \(\mathrm{kg} )\) wins a race with a skateboarder. The boy has speed 1.40 \(\mathrm{m} / \mathrm{s}\) at the crest of a slope 2.60 \(\mathrm{m}\) high and 12.4 \(\mathrm{m}\) long. At the bottom of the slope his speed is 6.20 \(\mathrm{m} / \mathrm{s}\) . If air resistance and rolling resistance can be modeled as a constant friction force of 41.0 \(\mathrm{N}\) , find the work he did in pushing forward on his wheels during the downhill ride.

A single conservative force acts on a \(5.00-\mathrm{kg}\) particle. The equation \(F_{x}=(2 x+4) \mathrm{N}\) describes the force, where \(x\) is in meters. As the particle moves along the \(x\) axis from \(x=1.00 \mathrm{m}\) to \(x=5.00 \mathrm{m},\) calculate (a) the work done by this force, (b) the change in the potential energy of the system, and \((c)\) the kinetic energy of the particle at \(x=5.00 \mathrm{m}\) if its speed is 3.00 \(\mathrm{m} / \mathrm{s}\) at \(x=1.00 \mathrm{m} .\)

A 1.00 -kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.60). The object has a speed of \(v_{i}=3.00 \mathrm{m} / \mathrm{s}\) when it makes contact with a light spring that has a force constant of 50.0 \(\mathrm{N} / \mathrm{m}\) . The object comes to rest after the spring has been compressed a distance \(d .\) The object is then forced toward the left by the spring and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance \(D\) to the left of the unstretched spring. Find (a) the distance of compression \(d,\) (b) the speed \(v\) at the unstretched position when the object is moving to the left, and \((\mathrm{c})\) the distance \(D\) where the object comes to rest.

The potential energy function for a system is given by \(U(x)=-x^{3}+2 x^{2}+3 x\) (a) Determine the force \(F_{x}\) as a function of \(x\) . (b) For what values of \(x\) is the force equal to zero? (c) Plot \(U(x)\) versus \(x\) and \(F_{x}\) versus \(x\) , and indicate points of stable and unstable equilibrium.
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