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Chapter 9: Problem 17

Luge Rider A luge and rider, with a total mass of \(85 \mathrm{~kg}\), emerge from a downhill track onto a horizontal straight track with an initial speed of \(37 \mathrm{~m} / \mathrm{s}\). If they slow at a constant rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) (a) what magnitude \(F\) is required for the slowing force, (b) what distance \(d\) do they travel while slowing, and (c) what work \(W\) is done on them by the slowing force? What are (d) \(F,(\mathrm{e}) d\), and (f) \(W\) if the luge and the rider slow at a rate of \(4.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
These are: (a) 170 N, (b) 342.25 m, (c) 58182.5 J, (d) 340 N, (e) 171.125 m, (f) 58182.5 J.

Step by step solution

01

Title - Find the Required Force (2.0 m/s虏)

To find the magnitude of the force required to slow the luge and rider, use Newton's second law: \( F = ma \) Here, \( m = 85 \mathrm{~kg} \) (mass of luge and rider), \( a = 2.0 \mathrm{~m/s}^2 \) (acceleration). Substitute these values in: \[ F = 85 \mathrm{~kg} \times 2.0 \mathrm{~m/s}^2 \] \[ F = 170 \mathrm{~N} \]
02

Title - Determine Distance Traveled (2.0 m/s虏)

To find the distance traveled while slowing, use the kinematic equation: \( v^2 = u^2 + 2ad \) Here, \( v = 0 \mathrm{~m/s} \) (final velocity), \( u = 37 \mathrm{~m/s} \) (initial velocity), \( a = -2.0 \mathrm{~m/s}^2 \) (negative because it is slowing), and solve for \( d \). Rearrange the equation to solve for \( d \): \[ 0 = (37 \mathrm{~m/s})^2 + 2 \times (-2.0 \mathrm{~m/s}^2) \times d \] \[ 0 = 1369 - 4d \] \[ 4d = 1369 \] \[ d = \frac{1369}{4} \approx 342.25 \mathrm{~m} \]
03

Title - Calculate Work Done (2.0 m/s虏)

Work done by the slowing force can be calculated using the work-energy principle: \( W = Fd \) Here, \( F = 170 \mathrm{~N} \), and \( d = 342.25 \mathrm{~m} \). Substitute these values in: \[ W = 170 \mathrm{~N} \times 342.25 \mathrm{~m} \] \[ W = 58182.5 \mathrm{~J} \]
04

Title - Find the Required Force (4.0 m/s虏)

For the second scenario with a slowing rate of \( 4.0 \mathrm{~m/s}^2 \), use the same method as in Step 1: \( F = ma \) Here, \( m = 85 \mathrm{~kg} \), \( a = 4.0 \mathrm{~m/s}^2 \). Substitute these values in: \[ F = 85 \mathrm{~kg} \times 4.0 \mathrm{~m/s}^2 \] \[ F = 340 \mathrm{~N} \]
05

Title - Determine Distance Traveled (4.0 m/s虏)

For the second scenario, use the same kinematic equation as in Step 2: \( v^2 = u^2 + 2ad \) Here, \( v = 0 \mathrm{~m/s} \), \( u = 37 \mathrm{~m/s} \), and \( a = -4.0 \mathrm{~m/s}^2 \) (negative because it is slowing), and solve for \( d \). Rearrange the equation to solve for \( d \): \[ 0 = (37 \mathrm{~m/s})^2 + 2 \times (-4.0 \mathrm{~m/s}^2) \times d \] \[ 0 = 1369 - 8d \] \[ 8d = 1369 \] \[ d = \frac{1369}{8} \approx 171.125 \mathrm{~m} \]
06

Title - Calculate Work Done (4.0 m/s虏)

For the second scenario, use the work-energy principle as in Step 3: \( W = Fd \) Here, \( F = 340 \mathrm{~N} \), and \( d = 171.125 \mathrm{~m} \). Substitute these values in: \[ W = 340 \mathrm{~N} \times 171.125 \mathrm{~m} \] \[ W = 58182.5 \mathrm{~J} \]

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Key Concepts

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

kinematic equations
Kinematic equations are crucial in solving motion-related problems in physics. They describe the motion of an object under constant acceleration without considering the forces causing the motion. These equations are:
  • \[ v = u + at \]
  • \[ s = ut + \frac{1}{2}at^2 \]
  • \[ v^2 = u^2 + 2as \]
In these equations:
  • \( v \) is the final velocity
  • \( u \) is the initial velocity
  • \( a \) is the acceleration
  • \( t \) is the time
  • \( s \) is the displacement
In the exercise, we used the third equation:
\[ v^2 = u^2 + 2ad \]Given that the final velocity \( v \) was 0 m/s and using the initial velocity \( u \) and acceleration \( a \). We rearranged the equation to solve for the displacement \( d \). This equation is particularly helpful when time is not given directly.
work-energy principle
The work-energy principle states that the work done by all the forces acting on an object is equal to the change in its kinetic energy. Mathematically, the work done \( W \) is given by:
\[ W = Fd \] Here:
  • \( F \) is the force applied
  • \( d \) is the distance over which the force is applied
In the exercise, we calculated the work done by the slowing force as \( W = Fd \). We first found the force \( F \) by using Newton's Second Law:
\[ F = ma \]
Then we determined the distance \( d \) using the kinematic equation. This allowed us to calculate the work done as the product of force and distance. This principle is powerful because it ties the concepts of force and motion together, demonstrating how energy is transferred and transformed in mechanical processes.
constant acceleration
Constant acceleration occurs when an object's velocity changes by the same amount each second. This is a common scenario in many physics problems, such as the one involving the luge rider. When acceleration is constant, it simplifies the calculation of various motion parameters.
In the exercise, the acceleration \( a \) remained constant at \(-2.0 \, \mathrm{m/s^2} \) in the first scenario and \(-4.0 \, \mathrm{m/s^2} \) in the second scenario.
This constant rate of slowing made it possible to use kinematic equations directly to find distances traveled and work done. Constant acceleration allows for predictable analysis using straightforward equations, enabling efficient problem-solving in physics.

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Most popular questions from this chapter

A Sprinter A sprinter who weighs \(670 \mathrm{~N}\) runs the first \(7.0 \mathrm{~m}\) of a race in \(1.6 \mathrm{~s}\), starting from rest and accelerating uniformly. What are the sprinter's (a) speed and (b) kinetic energy at the end of the \(1.6 \mathrm{~s} ?\) (c) What average power does the sprinter generate during the 1.6 s interval?

Auto Starts from Rest A \(1500 \mathrm{~kg}\) automobile starts from rest on a horizontal road and gains a speed of \(72 \mathrm{~km} / \mathrm{h}\) in \(30 \mathrm{~s}\). (a) What is the kinetic energy of the auto at the end of the \(30 \mathrm{~s}\) ? (b) What is the average power required of the car during the \(30 \mathrm{~s}\) interval? (c) What is the instantaneous power at the end of the \(30 \mathrm{~s}\) interval, assuming that the acceleration is constant?

Large Meteorite vs. TNT On August 10,1972, a large meteorite skipped across the atmosphere above western United States and Canada, much like a stone skipped across water. The accompanying fireball was so bright that it could be seen in the daytime sky (see Fig. \(9-22\) for a similar event). The meteorite's mass was about \(4 \times 10^{6} \mathrm{~kg}\) : its speed was about \(15 \mathrm{~km} / \mathrm{s}\). Had it entered the atmosphere vertically, it would have hit Earth's surface with about the same speed. (a) Calculate the meteorite's loss of kinetic energy (in joules) that would have been associated with the vertical impact. (b) Express the energy as a multiple of the explosive energy of 1 megaton of \(\mathrm{TNT}\), which is \(4.2 \times 10^{15} \mathrm{~J}\). (c) The energy associated with the atomic bomb explosion over Hiroshima was equivalent to 13 kilotons of TNT. To how many Hiroshima bombs would the meteorite impact have been equivalent?

Floating Ice Block A floating ice block is pushed through a displacement \(\vec{d}=(15 \mathrm{~m}) \hat{\mathrm{i}}-(12 \mathrm{~m}) \hat{\mathrm{j}}\) along a straight embankment by rushing water, which exerts a force \(\vec{F}=(210 \mathrm{~N}) \mathrm{i}-(150 \mathrm{~N}) \hat{\mathrm{j}}\) on the block. How much work does the force do on the block during the displacement?

Lowering a Block A cord is used to vertically lower an initially stationary block of mass \(M\) at a constant downward acceleration of \(g / 4\). When the block has fallen a distance \(d\), find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block.
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