/*! 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 104 104 A four-person bobsled (total... [FREE SOLUTION] | 91Ó°ÊÓ

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104 A four-person bobsled (total mass \(=630 \mathrm{~kg}\) ) comes down a straightaway at the start of a bobsled run.The straightaway is \(80.0 \mathrm{~m}\) long and is inclined at a constant angle of \(10.2^{\circ}\) with the horizontal. Assume that the combined effects of friction and air drag produce on the bobsled a constant force of \(62.0 \mathrm{~N}\) that acts parallel to the incline and up the incline. Answer the following questions to three significant digits. (a) If the speed of the bobsled at the start of the run is \(6.20 \mathrm{~m} / \mathrm{s},\) how long does the bobsled take to come down the straightaway? (b) Suppose the crew is able to reduce the effects of friction and air drag to \(42.0 \mathrm{~N}\). For the same initial velocity, how long does the bobsled now take to come down the straightaway?

Short Answer

Expert verified
Time with 62 N drag: 9.66 s; with 42 N drag: 8.93 s.

Step by step solution

01

Determine the Net Force Acting on the Bobsled

To find the net force acting on the bobsled, start by calculating the gravitational force component along the incline: \[ F_g = mg \sin \theta \]where \( m = 630 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( \theta = 10.2^\circ \). Then, subtract the force of friction and air drag (62 N) to find the net force:\[ F_{\text{net}} = F_g - F_d \]
02

Calculate the Acceleration of the Bobsled

Using Newton's second law, calculate the acceleration of the bobsled:\[ F_{\text{net}} = ma \]Rearrange to solve for acceleration \( a \):\[ a = \frac{F_{\text{net}}}{m} \]
03

Use Kinematic Equation to Find the Time Taken

Applying the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:\[ v_f^2 = v_i^2 + 2a s \]Given that the final velocity \( v_f \) isn't needed, we rearrange to solve for time \( t \) using the equation:\[ s = v_i t + \frac{1}{2} a t^2 \]Solve this quadratic equation for \( t \) with initial velocity \( v_i = 6.20 \, \text{m/s} \) and displacement \( s = 80.0 \, \text{m} \).
04

Recalculate with Reduced Drag Force

Repeat the steps for calculating the net force, acceleration, and time by substituting the reduced drag force of 42 N in place of 62 N. This will give the new time taken.

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Key Concepts

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

Kinematics in Bobsled Dynamics
Kinematics is the branch of physics that deals with the motion of objects, without considering the forces that cause this motion. In the context of a bobsled moving down an inclined plane, kinematics allows us to determine how long it takes for the bobsled to travel a specific distance when starting from a certain speed.

The kinematic equation used here relates the initial velocity, acceleration, and distance to time. Since the bobsled has an initial speed of 6.20 m/s and must cover a straightaway of 80.0 meters, we can use \[ s = v_i t + \frac{1}{2} a t^2 \]This helps you solve for the time it takes to travel down the incline. Remember, this motion happens because of both initial speed and acceleration caused by gravitational pull and resistive forces.

Analyzing such equations aids in predicting how movements take place over time, which is crucial for understanding real-world dynamics of vehicles like bobsleds.
Newton's Second Law Applied
Newton's second law is a fundamental principle of physics that states that force equals mass times acceleration, or \[ F = ma \]In our bobsled scenario, this law helps to calculate acceleration based on the net force acting upon it. First, we identify the forces: the gravitational force component moving the bobsled downwards and the opposing friction and air resistance acting upwards.

By writing the net force equation as \[ F_{net} = F_g - F_d \]where \( F_d \) represents the drag forces, and solving for acceleration \[ a = \frac{F_{net}}{m} \],we can determine how quickly the bobsled speeds up or slows down. Once acceleration is known, it can be used in the kinematic equations to find time or other aspects of motion. Through Newton's second law, you can see how unbalanced forces directly affect an object's motion, a key aspect when designing or analyzing dynamic systems.
The Inclined Plane and Its Effects
An inclined plane is a flat surface tilted at an angle, rather than horizontal, which affects how gravity influences objects upon it. For the bobsled, this plane is inclined at a 10.2° angle, altering how gravity accelerates it down the slope.

The gravitational force component that affects the bobsled along the incline is \[ F_g = mg \sin \theta \]where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of incline. This equation helps determine how much of gravity's pull is effectively causing the bobsled to move downhill.

This component of the gravitational force needs to overcome resistive forces for the bobsled to accelerate. Understanding how inclined planes work highlights how angles can change the dynamics of movement and forces exerted on objects.
Influence of Friction and Air Resistance
Friction and air resistance are two forces that constantly act to slow down moving objects, such as a bobsled racing down an incline. These forces are generally described as resistive forces because they oppose the direction of motion.

In our scenario, these forces apply a constant counterforce of \( 62 \text{ N} \) (or a reduced \( 42 \text{ N} \) in the second part of exercise), opposing the bobsled's motion down the plane. Their impact can be seen when calculating net force using the formula:\[ F_{net} = F_g - F_d \]Where \( F_d \) is the drag force.

By understanding how changing the magnitude of friction and air resistance affects the net force, and consequently the acceleration, you can see how these factors dramatically influence performance. Reducing these resistive forces allows for quicker descent or better energy efficiency, which is essential for improving bobsledding performance.

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Most popular questions from this chapter

A \(2.5 \mathrm{~kg}\) block is initially at rest on a horizontal surface. A horizontal force \(\vec{F}\) of magnitude \(6.0 \mathrm{~N}\) and a vertical force \(\vec{P}\) are then applied to the block (Fig. \(6-17) .\) The coefficients of friction for the block and surface are \(\mu_{s}=0.40\) and \(\mu_{k}=0.25 .\) Determine the magnitude of the frictional force acting on the block if the magnitude of \(\vec{P}\) is (a) \(8.0 \mathrm{~N},\) (b) \(10 \mathrm{~N}\), and (c) \(12 \mathrm{~N}\).

An old streetcar rounds a flat corner of radius \(9.1 \mathrm{~m},\) at \(16 \mathrm{~km} / \mathrm{h} .\) What angle with the vertical will be made by the loosely hanging hand straps?

A locomotive accelerates a 25-car train along a level track. Every car has a mass of \(5.0 \times 10^{4} \mathrm{~kg}\) and is subject to a frictional force \(f=250 v,\) where the speed \(v\) is in meters per second and the force \(f\) is in newtons. At the instant when the speed of the train is \(30 \mathrm{~km} / \mathrm{h},\) the magnitude of its acceleration is \(0.20 \mathrm{~m} / \mathrm{s}^{2} .\) (a) \(\mathrm{What}\) is the tension in the coupling between the first car and the locomotive? (b) If this tension is equal to the maximum force the locomotive can exert on the train, what is the steepest grade up which the locomotive can pull the train at \(30 \mathrm{~km} / \mathrm{h} ?\)

As a \(40 \mathrm{~N}\) block slides down a plane that is inclined at \(25^{\circ}\) to the horizontal, its acceleration is \(0.80 \mathrm{~m} / \mathrm{s}^{2},\) directed up the plane. What is the coefficient of kinetic friction between the block and the plane?

An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed \(1100 \mathrm{~N}\). The coefficient of static friction between the box and the floor is \(0.35 .\) (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?

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