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An electric scooter has a battery capable of supplying 120 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 that the rider can achieve is calculated using the available energy and applying the principles of energy conservation. To get the right answer, it is important to correctly calculate the available energy after friction losses and appropriately apply the equation for gravitational potential energy.

Step by step solution

01

Calculate the Effective Energy

The energy consumed by the electric scooter is not fully used for propulsion as some gets lost to friction. However, we are given that \(60.0 \%\) of the energy is wasted due to losses. Thus, the effective useful energy \(E_{useful}\) from the scooter can be calculated as: \[E_{useful} = E_{total} - (E_{total} \times \%losses)\]This gives: \[E_{useful} = 120Wh - (120Wh \times 60.0\%)\]
02

Convert Energy to Joules

The result obtained in Wh has to be converted to Joules for further calculations. This conversion is done using the fact that 1Wh is equal to 3600 Joules.The energy \(E_{useful}\) can now be computed as: \[E_{useful} = 120Wh \times 3600 J/Wh \]
03

Calculate the Altitude Change

Using the principle of energy conservation, the scooter effectively converts the available energy into gravitational potential energy. The gravitational potential energy in this case can be calculated as \(P_{e}=mgh\). Thus, the altitude change which is 'h' can be found by rearranging the equation to \[h=\frac{E_{useful}}{mg}\]In this case, the weight \(W=mg=890 \, N\) and gravity \(g\), which is \(9.8 \, m/s^{2}\).Thus, the altitude change \(h\) in meters can now be computed as:\[h = \frac{E_{useful}}{mg}\]

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Key Concepts

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

Energy Conversion
Energy conversion is a fundamental concept in understanding how electric vehicles, like scooters, work. Essentially, it's about transforming one form of energy into another. In the case of an electric scooter, the battery provides electrical energy, which needs conversion to kinetic energy, motion, or in uphill scenarios, into potential energy for altitude change. However, most real-life energy systems are not 100% efficient.
In the given scenario, the energy conversion isn't perfect because a portion of energy is lost due to factors like friction. Here, 60% of the total energy from the battery is lost, meaning only 40% is effectively used for the scooter's primary purposes. This understanding of energy conversion highlights the importance of improving efficiency to make the most out of the available energy.
Remember:
  • Total energy from the battery must always account for losses you can't avoid (e.g., friction).
  • Energy lost is energy you cannot use for propulsion or climbing hills.
This principle applies broadly to all electric vehicles and is critical for achieving better energy efficiency.
Gravitational Potential Energy
Gravitational potential energy is the energy that an object possesses because of its height relative to the Earth. For an electric scooter carrying a rider uphill, this is the key form of energy that allows the scooter to climb.
The formula to calculate gravitational potential energy is given by \(P_{e} = mgh\), where:
  • \(m\) is the mass of the object (rider plus scooter weight in Newtons).
  • \(g\) is the acceleration due to gravity, which is approximately \(9.8 \text{ m/s}^2\) on Earth.
  • \(h\) is the height or altitude change, which we ultimately want to calculate.
Gravitational potential energy is vital because it influences how much work needs to be done to climb to a certain altitude. By understanding this concept, you can better calculate how battery energy translates into climbing potential.
Altitude Change Calculation
Calculation of altitude change using energy principles is a straightforward application of physics. Once you know the effective energy available from the scooter's battery, the next step is determining how high you can go with that energy.
After converting the available energy from watt-hours to joules (since 1 Wh = 3600 J), use the equation for gravitational potential energy \(E_{useful} = mgh\) to find the altitude change. Rearranging this formula, you find
\[h = \frac{E_{useful}}{mg}\].
  • \(h\) represents the altitude change in meters.
  • \(E_{useful}\) is effective energy in joules.
  • \(mg\) is the weight of the scooter plus rider, a constant at 890 N.
This calculation tells you how much height can be gained using the energy left after losses, grounding the concept of energy efficiency in tangible results like altitude change. By mastering these calculations, you get a clearer idea of the vehicle's capabilities under real conditions.

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Most popular questions from this chapter

Assume that you attend a state university that started out as an agricultural college. Close to the center of the campus is a tall silo topped with a hemispherical cap. The cap is frictionless when wet. Someone has somehow balanced a pumpkin at the highest point. The line from the center of curvature of the cap to the pumpkin makes an angle \(\theta_{i}=0^{\circ}\) with the vertical. While you happen to be standing nearby in the middle of a rainy night, a breath of wind makes the pumpkin start sliding downward from rest. It loses contact with the cap when the line from the center of the hemisphere to the pumpkin makes a certain angle with the vertical. What is this angle?

A \(5.00-\mathrm{kg}\) block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The other end of the spring is held fixed. The spring is compressed \(0.100 \mathrm{m}\) from equilibrium and released. The speed of the block is \(1.20 \mathrm{m} / \mathrm{s}\) when it passes the equilibrium position of the spring. The same experiment is now repeated with the frictionless surface replaced by a surface for which the coefficient of kinetic friction is \(0.300 .\) Determine the speed of the block at the equilibrium position of the spring.

The potential energy of a system of two particles separated by a distance \(r\) is given by \(U(r)=A / r,\) where \(A\) is a constant. Find the radial force \(\mathbf{F}_{r}\) that each particle exerts on the other.

The world's biggest locomotive is the MK5000C, a behemoth of mass 160 metric tons driven by the most powerful engine ever used for rail transportation, a Caterpillar diesel capable of 5000 hp. Such a huge machine can provide a gain in efficiency, but its large mass presents challenges as well. The engineer finds that the locomotive handles differently from conventional units, notably in braking and climbing hills. Consider the locomotive pulling no train, but traveling at \(27.0 \mathrm{m} / \mathrm{s}\) on a level track while operating with output power \(1000 \mathrm{hp} .\) It comes to a \(5.00 \%\) grade (a slope that rises \(5.00 \mathrm{m}\) for every \(100 \mathrm{m}\) along the track). If the throttle is not advanced, so that the power level is held steady, to what value will the speed drop? Assume that friction forces do not depend on the speed.

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.

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