/*! 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 35 A 55.0-kg skateboarder starts ou... [FREE SOLUTION] | 91Ó°ÊÓ

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A 55.0-kg skateboarder starts out with a speed of 1.80 m/s. He does 80.0 J of work on himself by pushing with his feet against the ground. In addition, friction does 265 J of work on him. In both cases, the forces doing the work are non conservative. The final speed of the skateboarder is 6.00 \(\mathrm{m} / \mathrm{s}\) . (a) Calculate the change \(\left(\Delta PE=PE_{\mathrm{f}}-PE_{0}\right)\) in the gravitational potential energy. (b) How much has the vertical height of the skater changed, and is the skater above or below the starting point?

Short Answer

Expert verified
The change in potential energy is -1085.9 J, and the skater is 2.01 m below the starting point.

Step by step solution

01

Identify the Type of Problem

This is a work-energy problem where changes in kinetic energy (KE), potential energy (PE), and the work done by the skateboarder and friction are involved. We need to find the change in potential energy and the change in vertical height.
02

Apply the Work-Energy Theorem

The work-energy theorem states that the work done by all forces is equal to the change in kinetic energy of the system: \( W_{\text{net}} = \Delta KE = KE_f - KE_0 \). We need to compute the net work done.
03

Calculate Initial and Final Kinetic Energy

Initial kinetic energy \( KE_0 \) is given by \( \frac{1}{2} m v_0^2 \) and final kinetic energy \( KE_f \) is \( \frac{1}{2} m v_f^2 \). For the skateboarder: \[ KE_0 = \frac{1}{2} \times 55.0 \, \text{kg} \times (1.80 \, \text{m/s})^2 = 89.1 \, \text{J} \] \[ KE_f = \frac{1}{2} \times 55.0 \, \text{kg} \times (6.00 \, \text{m/s})^2 = 990.0 \, \text{J} \]
04

Determine the Net Work Done

The net work \( W_{\text{net}} \) is the sum of work done by the skateboarder and the work done by friction: \[ W_{\text{net}} = W_{\text{skateboarder}} - W_{\text{friction}} = 80.0 \, \text{J} - 265.0 \, \text{J} = -185.0 \, \text{J} \]
05

Calculate the Change in Kinetic Energy

Using the work-energy theorem: \[ W_{\text{net}} = \Delta KE = KE_f - KE_0 = 990.0 \, \text{J} - 89.1 \, \text{J} = 900.9 \, \text{J} \]
06

Find the Change in Potential Energy

Since \( W_{\text{net}} = \Delta KE + \Delta PE \), we have: \[ -185.0 \, \text{J} = 900.9 \, \text{J} + \Delta PE \] Solving for \( \Delta PE \), \[ \Delta PE = -1085.9 \, \text{J} \]
07

Calculate the Change in Vertical Height

The change in potential energy \( \Delta PE \) is related to height \( h \) by \( \Delta PE = mgh \). Rearranging gives \[ h = \frac{\Delta PE}{mg} = \frac{-1085.9 \, \text{J}}{55.0 \, \text{kg} \times 9.81 \, \text{m/s}^2} = -2.01 \, \text{m} \].
08

Determine the Direction of Height Change

Since \( h = -2.01 \, \text{m} \), the vertical height has decreased, meaning the skater is below the starting point.

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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 the energy an object possesses due to its motion. For instance, when a skateboarder moves, they possess kinetic energy, which can be calculated using the formula: \( KE = \frac{1}{2} mv^2 \). Here, \( m \) is the mass, and \( v \) is the velocity of the object.
In practical scenarios like this exercise, knowing the skateboarder's speed helps in determining how much kinetic energy they have both at the beginning and the end of their motion.
  • The initial kinetic energy \( (KE_0) \) of the skateboarder was calculated with their starting speed.
  • By determining the final speed, we find the final kinetic energy \( (KE_f) \).
The change in kinetic energy is pivotal in analyzing how energy has been transferred or transformed during motion. Such changes give insight into the overall dynamics and help us solve real-world problems involving motion.
Potential Energy
Potential energy often relates to the height of an object relative to a specific point, usually the ground. In this context, gravitational potential energy is crucial. It is calculated using the formula: \( PE = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (approx. \( 9.81 \, \text{m/s}^2 \)), and \( h \) is the height.
This exercise demonstrated a change in the skateboarder's potential energy as they moved vertically.
  • The potential energy change \( (\Delta PE) \) reflects how the skateboarder's height has altered.
  • A negative change indicates that the skateboarder's vertical position is lower.
Analyzing how potential energy changes can help us understand the vertical movement or height change in various scenarios, such as when an object rises or falls. Understanding these principles is key to problem-solving in physics.
Non-Conservative Forces
Non-conservative forces are forces where the work done depends on the path taken and not just the initial and final states. Common examples include friction and air resistance. These forces can remove energy from the system or transfer it into other forms, like thermal energy.
In the context of this exercise, two non-conservative forces are at play:
  • The skateboarder exerted force by pushing against the ground, adding energy to the system.
  • Friction did work on the skateboarder, reducing the system's energy.
The net result of these non-conservative forces significantly impacts the calculations of kinetic and potential energy changes. In this particular scenario, understanding how these forces affect energy helps in pinning down why the skateboarder's height decreases rather than increases after the motion.

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Most popular questions from this chapter

A person pulls a toboggan for a distance of 35.0 m along the snow with a rope directed \(25.0^{\circ}\) above the snow. The tension in the rope is 94.0 N. (a) How much work is done on the toboggan by the tension force? (b) How much work is done if the same tension is directed parallel to the snow?

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