/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 A \(1250-\mathrm{kg}\) car drive... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(1250-\mathrm{kg}\) car drives up a hill that is \(16.2 \mathrm{m}\) high. During the drive, two non conservative forces do work on the car: (i) the force of friction, and (ii) the force generated by the car's engine. The work done by friction is \(-3.11 \times 10^{5} \mathrm{J} ;\) the work done by the engine is \(+6.44 \times 10^{5}\) J. Find the change in the car's kinetic energy from the bottom of the hill to the top of the hill.

Short Answer

Expert verified
The change in the car's kinetic energy is \(1.35 \times 10^5\, \text{J}\).

Step by step solution

01

Identify Known Values

We are given the following values:- Mass of the car: \( m = 1250\, \text{kg} \)- Height of the hill: \( h = 16.2\, \text{m} \)- Work done by friction: \( W_{\text{friction}} = -3.11 \times 10^5\, \text{J} \)- Work done by the engine: \( W_{\text{engine}} = +6.44 \times 10^5\, \text{J} \)
02

Calculate Gravitational Potential Energy Change

The change in gravitational potential energy \( \Delta U \) when the car moves up the hill is given by:\[ \Delta U = mgh \]Substitute the known values:\[ \Delta U = (1250\, \text{kg})(9.8\, \text{m/s}^2)(16.2\, \text{m}) \]\[ \Delta U = 1.98 \times 10^5\, \text{J} \]
03

Apply Work-Energy Principle

According to the work-energy principle, the total work done on the car is equal to the change in its kinetic energy \( \Delta KE \). The total work is the sum of the work done by all forces:\[ \Delta KE = W_{\text{net}} = W_{\text{engine}} + W_{\text{friction}} - \Delta U \]
04

Compute the Change in Kinetic Energy

Substitute the known values into the work-energy principle:\[ \Delta KE = 6.44 \times 10^5\, \text{J} + (-3.11 \times 10^5\, \text{J}) - 1.98 \times 10^5\, \text{J} \]Calculate:\[ \Delta KE = 6.44 \times 10^5 - 3.11 \times 10^5 - 1.98 \times 10^5 \]\[ \Delta KE = 1.35 \times 10^5\, \text{J} \]

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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 a crucial concept in understanding how objects move. It relates to the energy an object possesses due to its motion. The formula to calculate kinetic energy is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.
In the context of a car driving up a hill, this energy changes as the car moves, speeds up, or slows down. If the car speeds up, its kinetic energy increases. Conversely, if it slows down, its kinetic energy decreases. The change in kinetic energy is influenced by various factors, such as the work done by the engine and frictional forces.
Using the work-energy principle, we can determine how much kinetic energy changes by looking at the net work done by all forces acting on the car, which includes both positive work (like from the engine) and negative work (like from friction). Thus, understanding kinetic energy allows us to predict how the motion of the car will change as it ascends the hill.
Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses because of its position within a gravitational field. It depends on three key factors:
  • Mass of the object \( m \)
  • Gravitational acceleration \( g \) (approx. \( 9.8 \, \text{m/s}^2 \) on Earth)
  • Height \( h \) to which the object has been raised
If we consider a car driving up a hill, its gravitational potential energy increases as it climbs higher. The formula \( U = mgh \) helps calculate how much potential energy it has gained by reaching a certain height.
As given in the exercise, the car's potential energy changes by \( 1.98 \times 10^5 \, \text{J} \) as it ascends the 16.2-meter hill. This change is a crucial part of understanding the total energy transformations taking place as the car moves. By analyzing potential energy, we can better gauge how energy shifts between kinetic and potential states, helping to explain the work done by forces along the way.
Friction
Friction is a force that opposes the motion of objects sliding against each other. It's a non-conservative force because energy lost due to friction is not retrievable for mechanical work. Instead, it converts kinetic energy into thermal energy.
In the case of the car driving up the hill, friction acts against the forward motion, indicated by the negative work \( W_{\text{friction}} = -3.11 \times 10^5\, \text{J} \). This negative indicates that energy is removed from the system, effectively slowing the car down or increasing the work the engine must do.
Understanding friction helps us see why more power is needed to drive up a hill or why objects eventually come to a stop when no additional forces act on them. It plays a critical role in the real-world application of the Work-Energy Principle, illustrating how much energy is required just to overcome resistance to movement.

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Most popular questions from this chapter

At the local playground a child on a swing has a speed of \(2.02 \mathrm{m} / \mathrm{s}\) when the swing is at its lowest point. (a) To what maximum vertical height does the child rise, assuming he sits still and "coasts"? Ignore air resistance. (b) How do your results change if the initial speed of the child is halved?

A trapeze artist of mass \(m\) swings on a rope of length \(L\) Initially, the trapeze artist is at rest and the rope makes an angle \(\theta\) with the vertical. (a) Find the tension in the rope when it is vertical. (b) Explain why your result for part (a) depends on \(L\) in the way it does.

A \(1.9-\mathrm{kg}\) block slides down a frictionless ramp, as shown in Figure \(8-28 .\) The top of the ramp is \(1.5 \mathrm{m}\) above the ground; the bottom of the ramp is \(0.25 \mathrm{m}\) above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance \(d\) away. Find the distance \(d\).

Catching a wave, a 77 -kg surfer starts with a speed of \(1.3 \mathrm{m} / \mathrm{s}\) drops through a height of \(1.65 \mathrm{m},\) and ends with a speed of \(8.2 \mathrm{m} / \mathrm{s}\). How much nonconservative work was done on the surfer?

A 2.9-kg block slides with a speed of 1.6 \(\mathrm{m} / \mathrm{s}\) on a frictionless horizontal surface until it encounters a spring. (a) If the block compresses the spring \(4.8 \mathrm{cm}\) before coming to rest, what is the force constant of the spring? (b) What initial speed should the block have to compress the spring by \(1.2 \mathrm{cm} ?\)

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