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Chapter 6: Problem 89

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 \(\mathrm{m} / \mathrm{s}\) encounters a rough patch that reduces her speed to 1.65 \(\mathrm{m} / \mathrm{s}\) due to a friction force that is 25\(\%\) of her weight. Use the work-energy theorem to find the length of this rough patch.

Short Answer

Expert verified
The length of the rough patch is approximately 3.28 meters.

Step by step solution

01

Identify the Known Values

We start by jotting down the facts given in the problem. Initial speed of the skater, \( v_i = 3.0 \, \text{m/s} \); final speed, \( v_f = 1.65 \, \text{m/s} \); friction force is 25% of the skater's weight. The goal is to find the length of the rough patch, denoted as \( d \).
02

Apply the Work-Energy Theorem

The work-energy theorem states that the work done by the forces is equal to the change in kinetic energy. The work done by the friction force (\( F_f \)) is \( F_f \cdot d \), and the change in kinetic energy (\( \Delta KE \)) is \( \frac{1}{2}m {v_f}^2 - \frac{1}{2}m {v_i}^2 \).
03

Expression for Frictional Work Done

The work done by the friction force can be expressed as: \(-F_f \cdot d = \Delta KE \), where \( F_f = 0.25 \times mg \), with \( m \) being the skater's mass and \( g = 9.8 \, \text{m/s}^2 \).
04

Set Up the Equation

Substitute \( F_f = 0.25mg \) into the equation: \(-0.25mg \cdot d = \frac{1}{2}m(1.65^2) - \frac{1}{2}m(3.0^2) \). Notice that the mass \( m \) cancels out from the equation.
05

Simplify and Solve for \( d \)

Calculate the change in kinetic energy: \( \Delta KE = \frac{1}{2}(1.65^2) - \frac{1}{2}(3.0^2) \). Evaluate this value, then solve the equation for \( d \). The equation becomes \(-0.25g \cdot d = \Delta KE \), leading to \( d = \frac{-\Delta KE}{0.25g} \).
06

Calculate the Length of the Rough Patch

Compute \( d = \frac{-\Delta KE}{0.25 \times 9.8} \). First, determine \( \Delta KE = \frac{1}{2}(1.65^2 - 3.0^2) \) which equals approximately \(-8.0475 \). Then compute \( d = \frac{8.0475}{0.25 \times 9.8} \), which is approximately 3.28 meters.

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Key Concepts

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

Friction Force
Friction is a resistive force that acts to oppose the direction of motion of an object. It is due to the roughness of the surfaces in contact and can significantly affect motion. In the exercise, the skater experiences a friction force that is 25% of her weight when she glides over a rough patch on the ice rink. This friction acts against her motion, causing her to slow down.

The formula to determine this friction force is given by:
  • Friction force ( F_f ) = 0.25 imes mg
Where mg represents the skater's weight, with m being her mass and g being the acceleration due to gravity (approximated as 9.8 m/s虏).

This indicates that friction is directly proportional to weight. So, the heavier the skater, the larger the friction force she will encounter. Understanding friction is crucial as it converts kinetic energy into heat or sound, slowing down the skater on her path.
Kinetic Energy
Kinetic energy refers to the energy that an object possesses due to its motion. It is dependent on two factors: the mass of the object and the square of its velocity.

Kinetic energy is calculated using the formula:
  • Kinetic energy (KE) = \( \frac{1}{2}m v^2 \)
Where m is the mass of the object, and v is its velocity.

In the context of the skater, her initial and final kinetic energies are critical in determining how much energy has been lost due to friction:
  • Initial kinetic energy: \(\frac{1}{2}mv_i^2\)
  • Final kinetic energy: \(\frac{1}{2}mv_f^2\)
The change in kinetic energy (螖KE) helps us evaluate the energy transferred from motion into overcoming friction, which ultimately determines how far the skater travels along the rough patch. This calculation is integral to applying the work-energy theorem, linking the work done by friction to the change in kinetic energy.
Horizontal Motion
Horizontal motion refers to the movement along a straight line parallel to the horizon. In this exercise, the skater's horizontal motion takes place on an ice rink, where factors like friction come into play significantly to alter her speed.

In horizontal motion, the work-energy theorem comes into effect, especially when unbalanced forces like friction act on an object. This theorem ties together the work done by the forces to the change in the object's kinetic energy.
  • The theorem stipulates that: Work done = Change in kinetic energy ( 螖KE )
The skater's motion is impeded by the friction force that acts on her. This interaction exemplifies how even in the absence of vertical forces (since it's horizontal), friction can slow down the movement. The work done by friction extracts energy from the skater鈥檚 motion, reducing her speed until she comes to a smoother surface.

This concept highlights the interconnected nature of physics concepts鈥攗nderstanding how they relate is key to deciphering problems involving motion and forces on a horizontal plane.

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Most popular questions from this chapter

Automotive Power II. (a) If 8.00 hp are required to drive a \(1800-\) -kg automobile at 60.0 \(\mathrm{km} / \mathrm{h}\) on a level road, what is the total retarding force due to friction, air resistance, and so on? (b) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) up a 10.0\(\%\) grade (a hill rising 10.0 \(\mathrm{m}\) vertically in 100.0 \(\mathrm{m}\) horizon- tally)? (c) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) down a 1.00\(\%\) grade? (d) Down what percent grade would the car coast at 60.0 \(\mathrm{km} / \mathrm{h}\) ?

At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large compressed spring. The spring with force constant \(k=40.0 \mathrm{N} / \mathrm{cm}\) and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 70.0 \(\mathrm{kg}\) are pushed against the other end, compressing the spring 0.375 \(\mathrm{m}\) . The sled is then released with zero initial velocity. What is the sled's speed when the spring (a) returns to its uncompressed length and (b) is still compressed 0.200 \(\mathrm{m} ?\)

CALC An object is attracted toward the origin with a force given by \(F_{x}=-k / x^{2}\) . (Gravitational and electrical forces have this distance dependence.) (a) Calculate the work done by the force \(F_{x}\) when the object moves in the \(x\) -direction from \(x_{1}\) to \(x_{2}\) . If \(x_{2}>x_{1},\) is the work done by \(F_{x}\) positive or negative? (b) The only other force acting on the object is a force that you exert with your hand to move the object slowly from \(x_{1}\) to \(x_{2} .\) How much work do you do? If \(x_{2}>x_{1},\) is the work you do positive or negative? (c) Explain the similarities and differences between your answers to parts (a) and (b).

Six diesel units in series can provide 13.4 \(\mathrm{MW}\) of power to the lead car of a freight train. The diesel units have total mass \(1.10 \times 10^{6} \mathrm{kg}\) . The average car in the train has mass \(8.2 \times 10^{4} \mathrm{kg}\) and requires a horizontal pull of 2.8 \(\mathrm{kN}\) to move at a constant 27 \(\mathrm{m} / \mathrm{s}\) on level tracks. (a) How many cars can be in the train under these conditions? (b) This would leave no power for accelerating or climbing hills. Show that the extra force needed to accelerate the train is about the same for a \(0.10-\mathrm{m} / \mathrm{s}^{2}\) acceleration or a 1.0\(\%\) slope (slope angle \(\alpha=\arctan 0.010 )\) . (c) With the 1.0\(\%\) slope, show that an extra 2.9 \(\mathrm{MW}\) of power is needed to maintain the \(27-\mathrm{m} / \mathrm{s}\) speed of the diesel units. (d) With 2.9 \(\mathrm{MW}\) less power available, how many cars can the six diesel units pull up a 1.0\(\%\) slope at a constant 27 \(\mathrm{m} / \mathrm{s} ?\)

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \(\vec{F}=(30 \mathrm{N}) \hat{\imath}-(40 \mathrm{N}) \hat{\mathrm{J}}\) to the cart as it undergoes a displacement \(\vec{s}=(-9.0 \mathrm{m}) \hat{\boldsymbol{\imath}}-(3.0 \mathrm{m}) \hat{\boldsymbol{J}}\) . How much work does the force you apply do on the grocery cart?
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