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Chapter 8: Problem 109

A \(60.0 \mathrm{~kg}\) circus performer slides \(4.00 \mathrm{~m}\) down a pole to the circus floor, starting from rest. What is the kinetic energy of the performer as she reaches the floor if the frictional force on her from the pole (a) is negligible (she will be hurt) and (b) has a magnitude of \(500 \mathrm{~N}\) ?

Short Answer

Expert verified
(a) KE = 2352 J, (b) KE = 352 J

Step by step solution

01

Understand the Problem

To solve this we need to calculate the kinetic energy when the performer reaches the ground. To do this, we use the principle of work-energy which states that the work done on the performer will equal her change in kinetic energy.
02

Calculate Gravitational Force

The gravitational force acting on the performer is given by \( F_{gravity} = m \cdot g \), where \( m = 60.0 \mathrm{~kg} \) is the mass, and \( g = 9.8 \mathrm{~m/s^2} \) is the acceleration due to gravity. Thus, \( F_{gravity} = 60.0 \cdot 9.8 = 588 \mathrm{~N} \).
03

Compute Work Done by Gravity

The work done by gravity as she slides down the pole is \( W = F \cdot d \), where \( F = 588 \mathrm{~N} \) and \( d = 4.00 \mathrm{~m} \). Therefore, \( W = 588 \cdot 4 = 2352 \mathrm{~J} \).
04

Evaluate Kinetic Energy for Part (a)

Since the frictional force is negligible, all the work done by gravity is converted into kinetic energy. Thus, the kinetic energy \( KE = 2352 \mathrm{~J} \).
05

Compute Work Done Against Friction for Part (b)

With a frictional force of \( 500 \mathrm{~N} \), the work done against friction is \( W_{friction} = F_{friction} \cdot d = 500 \cdot 4 = 2000 \mathrm{~J} \).
06

Evaluate Kinetic Energy for Part (b)

The net work done on the performer is \( W_{net} = W - W_{friction} = 2352 - 2000 = 352 \mathrm{~J} \). Hence, the kinetic energy in this case is \( 352 \mathrm{~J} \).

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in physics that connects the work done on an object to its change in kinetic energy.
Essentially, when work is done on an object, it results in a change in that object's energy.
  • The total work done on the object by external forces changes its kinetic energy.
  • In mathematical terms, it is stated as \[ \Delta KE = W_{total}, \]where \( \Delta KE \) is the change in kinetic energy and \( W_{total} \) is the total work done.
In the circus performer problem, we see this principle in action. As the performer slides down the pole, the work done by gravitational forces converts into kinetic energy. Initially, she starts at rest, implying her starting kinetic energy is zero. As she slides down, the work-energy principle states that her kinetic energy when reaching the floor equals the total work done on her.
Frictional Force
Frictional force is a resistive force that acts opposite to the direction of motion when two surfaces are in contact.
In simpler terms, it tries to slow down moving objects.
  • Friction arises due to the microscopic interactions between the surfaces.
  • Its magnitude can vary greatly depending on surface texture, materials, and other factors.
In the given problem, the performer experiences a frictional force of magnitude \(500 \mathrm{~N}\). If this frictional force is significant, it transforms some of the gravitational work into overcoming friction, thus reducing the kinetic energy gained during her slide. This example explicitly demonstrates how friction can affect an object's motion and resultant energy.
Gravitational Force
Gravitational force is an attractive force exerted by Earth on any object with mass, pulling it towards the planet's center.
It is determined by the formula:
  • \[ F_{gravity} = m \cdot g, \]where \( m \) is the mass of the object, and \( g = 9.8 \mathrm{~m/s^2} \) is the acceleration due to gravity.
In the context of our exercise, the gravitational force on the circus performer is calculated as \(588 \mathrm{~N}\) due to her mass of \(60.0 \mathrm{~kg}\). This force is responsible for doing the work that increases her kinetic energy as she slides down the pole. Without friction, all of this force's work contributes to her speed when reaching the floor, showing gravitational force's direct impact on kinetic energy changes.
Work Done Against Friction
The work done against friction refers to the energy used in overcoming the frictional force when an object moves.
This energy is not transferred to the object's kinetic energy but is instead dissipated, often as heat.
  • The calculation for work done against friction is:\[ W_{friction} = F_{friction} \cdot d, \]where \( F_{friction} \) is the frictional force's magnitude, and \( d \) is the distance over which it acts.
  • This work directly reduces the net work available to transform into kinetic energy.
Considering the performer's situation with a frictional force of \(500 \mathrm{~N}\), the work done against friction totals \(2000 \mathrm{~J}\). This value represents the energy diverted to overcome friction during her descent, thereby lowering the kinetic energy she gains relative to a frictionless slide. This illustrates the energy trade-offs inherent when friction is present.

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Most popular questions from this chapter

A \(3.5 \mathrm{~kg}\) block is accelerated from rest by a compressed spring of spring constant \(640 \mathrm{~N} / \mathrm{m}\). The block leaves the spring at the spring's relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction \(\mu_{k}=0.25 .\) The frictional force stops the block in distance \(D=7.8 \mathrm{~m}\). What are (a) the increase in the thermal energy of the block-floor system, (b) the maximum kinetic energy of the block, and (c) the original compression distance of the spring?

You drop a \(2.00 \mathrm{~kg}\) book to a friend who stands on the ground at distance \(D=10.0 \mathrm{~m}\) below. If your friend's outstretched hands are at distance \(d=\) \(1.50 \mathrm{~m}\) above the ground (Fig. \(8-28)\), (a) how much work \(W_{g}\) does the gravitational force do on the book as it drops to her hands? (b) What is the change \(\Delta U\) in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy \(U\) of that system is taken to be zero at ground level, what is \(U(\mathrm{c})\) when the book is released and (d) when it reaches her hands? Now take \(U\) to be \(100 \mathrm{~J}\) at ground level and again find (e) \(W_{g},(\mathrm{f})\) \(\Delta U,(\mathrm{~g}) U\) at the release point, and (h) \(U\) at her hands.

A small block is sent through point \(A\) with a speed of \(7.0 \mathrm{~m} / \mathrm{s}\). Its path is without friction until it reaches the section of length \(L=12 \mathrm{~m}\), where the coefficient of kinetic friction is \(0.70 .\) The indicated heights are \(h_{1}=6.0 \mathrm{~m}\) and \(h_{2}=2.0 \mathrm{~m} .\) What are the speeds of the block at (a) point \(B\) and (b) point \(C ?\) (c) Does the block reach point \(D ?\) If so, what is its speed there; if not, how far through the section of friction does it travel?

The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$U=\frac{A}{r^{12}}-\frac{B}{r^{6}}$$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation - that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

A \(50 \mathrm{~g}\) ball is thrown from a window with an initial velocity of \(8.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) above the horizontal. Using energy methods, determine (a) the kinetic energy of the ball at the top of its flight and (b) its speed when it is \(3.0 \mathrm{~m}\) below the window. Does the answer to (b) depend on either (c) the mass of the ball or (d) the initial angle?
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