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Chapter 7: Problem 13

A luge and its 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 a force slows them to a stop at a constant rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) (a) what magnitude \(F\) is required for the force, (b) what distance \(d\) do they travel while slowing, and (c) what work \(W\) is done on them by the force? What are (d) \(F,(\mathrm{e}) d,\) and \((\mathrm{f}) \mathrm{W}\) if they, instead, slow at \(4.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
For the initial deceleration, force is 170 N, distance 342.25 m, work 58182.5 J. For the second scenario, force is 340 N, distance 171.125 m, work 58182.5 J.

Step by step solution

01

Calculate the Required Force (a,d)

The force required to bring the luge and rider to a stop can be calculated using Newton's second law, \( F = ma \) where \( m = 85 \text{ kg} \) and \( a = -2.0 \text{ m/s}^2 \). Since the force is acting opposite to the direction of motion, \[F = 85 \times (-2.0) = -170 \text{ N}\]The magnitude of the force is \( |F| = 170 \text{ N}\).
02

Calculate the Distance Traveled (b,e)

To find the distance \( d \) traveled while slowing down, use the equation of motion \( v^2 = u^2 + 2ad \), where \( v = 0 \text{ m/s} \) (final speed), \( u = 37 \text{ m/s} \) (initial speed), and \( a = -2.0 \text{ m/s}^2 \).\[0 = 37^2 + 2(-2.0)d\]\[d = \frac{-37^2}{2(-2.0)} = \frac{1369}{4} = 342.25 \text{ meters}\]
03

Calculate the Work Done (c,f)

Work done \( W \) is calculated using \( W = F \times d \). With \( |F| = 170 \text{ N} \) and \( d = 342.25 \text{ m} \);\[W = 170 \times 342.25 = 58182.5 \text{ J}\]
04

Re-calculate Force for New Deceleration (d)

If the deceleration is \(4.0 \text{ m/s}^2\), then \[F = 85 \times (-4.0) = -340 \text{ N}\]The magnitude of the force is \( |F| = 340 \text{ N} \).
05

Re-calculate Distance for New Deceleration (e)

With the new deceleration \( a = -4.0 \text{ m/s}^2 \), use the equation of motion:\[0 = 37^2 + 2(-4.0)d\]\[d = \frac{-37^2}{2(-4.0)} = \frac{1369}{8} = 171.125 \text{ meters}\]
06

Re-calculate the Work Done for New Deceleration (f)

If \( |F| = 340 \text{ N} \) and \( d = 171.125 \text{ m} \), then work done \( W \) is:\[W = 340 \times 171.125 = 58182.5 \text{ J}\]

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Key Concepts

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

Deceleration
Deceleration is simply negative acceleration, meaning a decrease in speed over time. In physics, this happens when an object slows down due to a force acting opposite to the direction of motion. For instance, when a luge and rider emerge from a downhill track onto a flat one, they won't just stop without some force causing deceleration. Here, the deceleration values given are \(-2.0 \text{ m/s}^2\) and \(-4.0 \text{ m/s}^2\). The negative sign indicates that it's a deceleration, slowing down the rider with each second that passes.
  • The force causing deceleration is crucial because it determines how quickly an object comes to a stop. This ties directly into Newton's Second Law, \(F = ma\), where \(F\) is the force, \(m\) is the mass, and \(a\) is the acceleration.
  • Understanding deceleration helps in calculating other quantities like the stopping distance, which depends on how quickly the object decelerates.
Work-Energy Principle
The work-energy principle is an important concept that ties together work done by forces and the energy changes they cause. In this problem, as the luge and rider come to a stop, the work done by the decelerating force brings their kinetic energy down to zero. The principle states that the work done on an object is equal to the change in its kinetic energy.
  • Initial kinetic energy of the luge is given by \( \frac{1}{2} m v^2 \), where \(m = 85 \text{ kg}\) and \(v = 37 \text{ m/s}\).
  • The work \(W\) is calculated as the product of the force applied and the distance over which it acts: \(W = F \times d\).
  • When a force acts over a distance to bring the luge to rest, this work is negative because it removes energy from the luge's motion.
  • This principle helps us calculate how far the luge travels while stopping, a necessary value for understanding the motion involved.
Equations of Motion
Equations of motion are mathematical formulations that describe the motion of an object under the influence of forces. These equations allow us to find unknown values such as the stopping distance or the final velocity of an object. For a decelerating luge, the equation used is:
  • \(v^2 = u^2 + 2ad\): Here, \(v\) is the final velocity (0 \text{ m/s}), \(u\) is the initial velocity (37 \text{ m/s}), \(a\) is acceleration, and \(d\) is the distance traveled.
  • Setting \(v = 0\), we rearrange the equation to solve for \(d\), showing how initially fast-moving objects require a certain distance to come to a stop.
  • These equations are essential tools in solving motion-related problems, providing a clear mathematical pathway to follow.
  • Whether the deceleration is \(-2.0 \text{ m/s}^2\) or \(-4.0 \text{ m/s}^2\), these formulas allow us to calculate how quickly and over what distance the luge will stop, following Newton's laws.

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Most popular questions from this chapter

On August \(10,1972,\) a large meteorite skipped across the atmosphere above the western United States and western Canada, much like a stone skipped across water. The accompanying fireball was so bright that it could be seen in the daytime sky and was brighter than the usual meteorite trail. 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?

A \(12.0 \mathrm{~N}\) force with a fixed orientation does work on a particle as the particle moves through the three-dimensional displacement \(\vec{d}=(2.00 \hat{i}-4.00 \mathrm{j}+3.00 \mathrm{k}) \mathrm{m} .\) What is the angle between the force and the displacement if the change in the particle's kinetic energy is (a) \(+30.0 \mathrm{~J}\) and (b) \(-30.0 \mathrm{~J} ?\)

A \(4.00 \mathrm{~kg}\) block is pulled up a frictionless inclined plane by a \(50.0 \mathrm{~N}\) force that is parallel to the plane, starting from rest. The normal force on the block from the plane has magnitude \(13.41 \mathrm{~N}\). What is the block's speed when its displacement up the ramp is \(3.00 \mathrm{~m} ?\)

A can of sardines is made to move along an \(x\) axis from \(x=0.25 \mathrm{~m}\) to \(x=1.25 \mathrm{~m}\) by a force with a magnitude given by \(F=\exp \left(-4 x^{2}\right),\) with \(x\) in meters and \(F\) in newtons. (Here exp is the exponential function.) How much work is done on the can by the force?

A \(0.30 \mathrm{~kg}\) ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring \((k=500 \mathrm{~N} / \mathrm{m})\) whose other end is fixed. The ladle has a kinetic energy of \(10 \mathrm{~J}\) as it passes through its equilibrium position (the point at which the spring force is zero). (a) At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? (b) At what rate is the spring doing work on the ladle when the spring is compressed \(0.10 \mathrm{~m}\) and the ladle is moving away from the equilibrium position?
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