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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

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Question: Calculate the total work done by a runner with a mass of 83.0 kg as they run 400m and go over two hills with heights of 50m and 28m respectively. The runner starts with an initial speed of 6.50 m/s and ends with a final speed of 4.50 m/s while experiencing a resistance force of 9.00 N throughout the run. Answer: To find the total work done by the runner, we should first calculate the potential energy change, kinetic energy change, and work done against the resistive force. Then, we can apply the work-energy theorem to find the total work done.

Step by step solution

01

Calculate the potential energy change

To calculate the change in potential energy (ΔPE), we need to find the height difference between the two hills and use the following formula: ΔPE = m × g × Δh Where m is the runner's mass, g is the acceleration due to gravity (approximately 9.81 m/s²), and Δh is the height difference between the two hills. We are given that the height difference is 50m - 28m, so Δh = 22m. Therefore, we can calculate ΔPE as: ΔPE = (83.0kg) × (9.81m/s²) × (22m)
02

Calculate the kinetic energy change

To find the change in kinetic energy (ΔKE), we use the following formula: ΔKE = (1/2) × m × (v_final² - v_initial²) Where m is the runner's mass, v_initial is the initial speed, and v_final is the final speed. Given the initial speed of 6.50 m/s and the final speed of 4.50 m/s, we can calculate ΔKE as: ΔKE = (1/2) × (83.0kg) × [(4.50m/s)² - (6.50m/s)²]
03

Calculate the work done against the resistive force

Given the constant resistance force of 9.00 N, we can calculate the work done against this force (W_resistance) using the formula: W_resistance = F_resistance × distance The total distance the runner covers is 400m, so the work done against the resistance is: W_resistance = (9.00 N) × (400 m)
04

Apply the work-energy theorem

Now we will use the work-energy theorem, which states that the total work done (W) is equal to the change in kinetic energy (ΔKE) plus the change in potential energy (ΔPE) plus the work done against resistance (W_resistance): W = ΔKE + ΔPE + W_resistance We can plug in the values we have calculated in the previous steps to find the total work done by the runner: W = ΔKE + ΔPE + W_resistance After calculating the values for each term, we will be able to find the work done by the runner over the total distance.

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Key Concepts

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

Potential Energy Change
Imagine you need to move a box to the top shelf in your room — you have to put in effort to lift it against gravity. This is similar to what potential energy is all about. In physics, potential energy change refers to the energy stored in an object due to its position relative to a certain level, such as the ground. For objects near Earth's surface, this energy can be calculated using the formula

\( \text{{ΔPE}} = m \times g \times \text{{Δh}} \).

Here, 'm' stands for mass, 'g' for the acceleration due to gravity (which is approximately 9.81 m/s² on Earth), and 'Δh' for the change in height. In our example, a runner gaining or losing height while traversing hills is doing work against or with the force of gravity, resulting in a change in potential energy.
  • Calculating the change in potential energy helps us understand how much energy is stored or released as the runner moves up or down.
  • This concept plays a crucial role in the work-energy theorem when analyzing the runner's motion.

When the runner ascends or descends hills, the potential energy increases with elevation and decreases when coming down, respectively. By calculating the potential energy difference between the top of two different hills, you can find out how much energy the runner has gained or lost just due to elevation changes.
Kinetic Energy Change
Now let's focus on the energy associated with motion. Kinetic energy change occurs when an object accelerates or decelerates, altering its velocity. In the context of our runner, kinetic energy is tied to how fast he is moving. The formula to find the change in kinetic energy (ΔKE) is

\( \text{{ΔKE}} = \frac{1}{2} \times m \times (\text{{v_final}}^2 - \text{{v_initial}}^2) \).

'm' is the mass of the runner, 'v_initial' his initial velocity, and 'v_final' his velocity at a different time. If he speeds up, kinetic energy increases, and if he slows down, kinetic energy decreases.
  • Knowing the initial and final speeds allows you to calculate how much kinetic energy has been gained or lost during his run.
  • The runner's kinetic energy is highest at his maximum speed and lowest when he stops.

Figuring out the kinetic energy change helps us understand how the runner's speed fluctuations impact his overall energy usage, which forms an integral part of understanding the total work done in any physical activity.
Resistive Force Work
Imagine running against a strong wind or moving a sled through sand; these scenarios involve a resistance force that opposes motion. In physics, resistive force work refers to the work done to overcome forces like friction or air resistance that oppose movement. For the runner, even on a calm day, he faces a resistive force which could be due to the road's friction or drag from the air. The work done against this resistive force over a certain distance is given by

\( W_{\text{resistance}} = F_{\text{resistance}} \times \text{distance} \).

'F_resistance' is the resistive force and 'distance' is how far the runner travels while overcoming this force.
  • This formula calculates the energy the runner expends to keep moving despite resistance.
  • In our scenario, a constant resistive force means the runner consistently uses energy to counteract it.

Considering this work is vital when computing the total work the runner has done, as it accounts for a significant part of the energy expenditure during his run.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that bridges the gap between work done and energy. It states that the work done by all forces acting on an object results in a change in the object's kinetic energy. In simpler words, when you do work on something, you're either giving it energy or taking energy away from it. This principle is neatly summed up by the equation

\( W = \text{{ΔKE}} + \text{{ΔPE}} + W_{\text{resistance}} \).

'W' is the total work done, 'ΔKE' the change in kinetic energy, 'ΔPE' the change in potential energy, and 'W_resistance' is the work done against resistive forces.
  • It helps us to understand the full picture of energy transformation during the runner's journey.
  • By applying this principle, you can also predict how different factors, like slope or wind, will affect the runner's performance.

For our exercising runner, using the work-energy principle allows us to calculate the total work he's done by combining his changes in speed, elevation, and the resistance he's faced. It is a powerful tool for solving problems where forces cause an object to move.

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Most popular questions from this chapter

A spring with \(k=10.0 \mathrm{~N} / \mathrm{cm}\) is initially stretched \(1.00 \mathrm{~cm}\) from its equilibrium length. a) How much more energy is needed to further stretch the spring to \(5.00 \mathrm{~cm}\) beyond its equilibrium length? b) From this new position, how much energy is needed to compress the spring to \(5.00 \mathrm{~cm}\) shorter than its equilibrium position?

A 1.00 -kg mass is suspended vertically from a spring with \(k=100 . \mathrm{N} / \mathrm{m}\) and oscillates with an amplitude of \(0.200 \mathrm{~m} .\) At the top of its oscillation, the mass is hit in such a way that it instantaneously moves down with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Determine a) its total mechanical energy, b) how fast it is moving as it crosses the equilibrium point, and c) its new amplitude.

You use your hand to stretch a spring to a displacement \(x\) from its equilibrium position and then slowly bring it back to that position. Which is true? a) The spring's \(\Delta U\) is positive. b) The spring's \(\Delta U\) is negative. c) The hand's \(\Delta U\) is positive. d) The hand's \(\Delta U\) is negative. e) None of the above statements is true.

A block of mass \(0.773 \mathrm{~kg}\) on a spring with spring constant \(239.5 \mathrm{~N} / \mathrm{m}\) oscillates vertically with amplitude \(0.551 \mathrm{~m}\). What is the speed of this block at a distance of \(0.331 \mathrm{~m}\) from the equilibrium position?

A 0.500 -kg mass is attached to a horizontal spring with \(k=100 . \mathrm{N} / \mathrm{m}\). The mass slides across a frictionless surface. The spring is stretched \(25.0 \mathrm{~cm}\) from equilibrium, and then the mass is released from rest. a) Find the mechanical energy of the system. b) Find the speed of the mass when it has moved \(5.00 \mathrm{~cm}\). c) Find the maximum speed of the mass.

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