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A runner reaches the top of a hill with a speed of \(6.50 \mathrm{~m} / \mathrm{s}\). He descends \(50.0 \mathrm{~m}\) and then ascends \(28.0 \mathrm{~m}\) to the top of the next hill. His speed is now \(4.50 \mathrm{~m} / \mathrm{s}\). The runner has a mass of \(83.0 \mathrm{~kg}\). The total distance that the runner covers is \(400 . \mathrm{m}\), and there is a constant resistance to motion of \(9.00 \mathrm{~N}\). Use energy considerations to find the work done by the runner over the total distance.

Short Answer

Expert verified
To find the total work done by the runner, we calculated the initial and final kinetic energies, the potential energy changes for the descent and ascent, the work done against constant resistance, and then applied the work-energy principle. The total work done by the runner was found to be approximately -21346 J (rounded to the nearest whole number).

Step by step solution

01

Calculate the initial and final kinetic energies

We can find the initial and final kinetic energies using the formula: \(KE = \frac{1}{2}mv^2\) Initial kinetic energy: \(KE_{initial} = \frac{1}{2}(83.0\,\text{kg})(6.50\,\text{m/s})^2 = 1757.125\,\text{J}\) Final kinetic energy: \(KE_{final} = \frac{1}{2}(83.0\,\text{kg})(4.50\,\text{m/s})^2 = 843.975\,\text{J}\)
02

Calculate the potential energy change

We can find the potential energy change using the formula: \(PE = mgh\) Potential energy change for descent: \(PE_{descent} = (83.0\,\text{kg})(9.81\,\text{m/s}^2)(-50.0\,\text{m}) = -40719.5\,\text{J}\) Potential energy change for ascent: \(PE_{ascent} = (83.0\,\text{kg})(9.81\,\text{m/s}^2)(28.0\,\text{m}) = 22885.64\,\text{J}\) Total potential energy change: \(PE_{total} = PE_{descent} + PE_{ascent} = -40719.5\,\text{J} + 22885.64\,\text{J} = -17833.86\,\text{J}\)
03

Calculate the work done against the constant resistance

The work done against the constant resistance can be calculated using the formula: \(W_{resistance} = -Fd\) We're given that the total distance covered by the runner is 400 meters, and the resistance force acting on the runner is 9.00 N. Plugging in the values: \(W_{resistance} = -(9.00\,\text{N})(400\,\text{m}) = -3600\,\text{J}\)
04

Apply the work-energy principle

According to the work-energy principle: \(W_{total} = \Delta KE + \Delta PE + W_{resistance}\) Plugging in the values calculated above: \(W_{total} = (843.975\,\text{J} - 1757.125\,\text{J}) + (-17833.86\,\text{J}) + (-3600\,\text{J}) = -21346.01\,\text{J}\) The work done by the runner over the total distance is approximately \(-21346\,\text{J}\).

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Key Concepts

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

Kinetic Energy Calculation
Kinetic energy represents the energy that an object possesses due to its motion. It can be calculated using the formula:
\[ KE = \frac{1}{2}mv^2 \]
Where:
  • \( m \) is the mass of the object in kilograms,
  • \( v \) is the velocity of the object in meters per second,
  • \( KE \) stands for kinetic energy in Joules.
This equation signifies that the kinetic energy is directly proportional to both the mass of the object and the square of its velocity. Thus, even a small increase in speed results in a significant rise in kinetic energy. In the example of the runner, the initial and final kinetic energies are crucial for understanding the total work done throughout the motion. As illustrated, the kinetic energy decreases from initial to final, which signifies a loss of energy likely due to the work done against resistance and change in potential energy.
Potential Energy Change
Potential energy is the energy stored in an object because of its position or configuration. For gravitational potential energy near the Earth's surface, the formula is:
\[ PE = mgh \]
Here:
  • \( m \) is the mass in kilograms,
  • \( g \) is the acceleration due to gravity \( (9.81\, \text{m/s}^2) \),
  • \( h \) is the height in meters,
  • \( PE \) stands for gravitational potential energy in Joules.
When the runner descends and ascends hills, the potential energy changes. Descending causes a loss of potential energy, indicated by a negative value, and ascending gains potential energy. The total potential energy change contributes to the understanding of energy exchange during the runner's activity. It is important to note that the change in potential energy depends only on the vertical position changed, not on the path taken, indicating a property of conservative forces.
Work Against Resistance
Work is the process of energy transfer when a force causes an object to move. The work done against a constant force, such as friction or air resistance, is calculated with the formula:
\[ W_{resistance} = -Fd \]
Where:
  • \( F \) is the constant force in Newtons,
  • \( d \) is the distance over which the force is applied in meters,
  • \( W_{resistance} \) denotes the work done against resistance, in Joules.
A negative sign indicates that work is done against the direction of the force. In our runner’s case, this resistance factor may include friction, wind resistance, and other forces opposing the motion. The negative value of the work done against resistance highlights that it takes energy away from the runner. This energy must be provided by the runner's metabolic processes, underscoring the additional effort required to overcome resistance in motion.

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Most popular questions from this chapter

A particle of mass \(m\) moving in the \(x y\) -plane is confined by a two- dimensional potential function, \(U(x, y)=\frac{1}{2} k\left(x^{2}+y^{2}\right)\). a) Derive an expression for the net force, \(\vec{F}=F_{x} \hat{x}+F_{y} \hat{y}\) b) Find the equilibrium point on the \(x y\) -plane. c) Describe qualitatively the effect of the net force. d) What is the magnitude of the net force on the particle at the coordinate (3.00,4.00) in centimeters if \(k=10.0 \mathrm{~N} / \mathrm{cm} ?\) e) What are the turning points if the particle has \(10.0 \mathrm{~J}\) of total mechanical energy?

An arrow is placed on a bow, the bowstring is pulled back, and the arrow is shot straight up into the air; the arrow then comes back down and sticks into the ground. Describe all of the changes in work and energy that occur.

A ball is thrown horizontally from the top of a building that is \(26.01 \mathrm{~m}\) high. The ball strikes the ground at an angle of \(41.86^{\circ}\) with respect to the horizontal. Neglecting air resistance, with what speed was the ball thrown?

Suppose you throw a 0.0520 -kg ball with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(30.0^{\circ}\) above the horizontal from a building \(12.0 \mathrm{~m}\) high. a) What will be its kinetic energy when it hits the ground? b) What will be its speed when it hits the ground?

A constant force of \(40.0 \mathrm{~N}\) is needed to keep a car traveling at constant speed as it moves \(5.00 \mathrm{~km}\) along a road. How much work is done? Is the work done on or by the car?

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