/*! 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 110 DATA A road heading due east pas... [FREE SOLUTION] | 91Ó°ÊÓ

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DATA A road heading due east passes over a small hill. You drive a car of mass \(m\) at constant speed \(v\) over the top of the hill, where the shape of the roadway is well approximated as an arc of a circle with radius \(R\). Sensors have been placed on the road surface there to measure the downward force that cars exert on the surface at various speeds. The table gives values of this force versus speed for your car: $$ \begin{array}{lcccccc} \text { Speed }(m / s) & 6.00 & 8.00 & 10.0 & 12.0 & 14.0 & 16.0 \\ \hline \text { Force }(\mathrm{N}) & 8100 & 7690 & 7050 & 6100 & 5200 & 4200 \end{array} $$ Treat the car as a particle. (a) Plot the values in such a way that they are well fitted by a straight line. You might need to raise the speed, the force, or both to some power. (b) Use your graph from part (a) to calculate \(m\) and \(R .\) (c) What maximum speed can the car have at the top of the hill and still not lose contact with the road?

Short Answer

Expert verified
After the steps above, the mass \(m\) of the car and radius \(R\) of the hill can be obtained from the graph's slope and y-intercept respectively. And the maximum speed at the top of the hill where the car still maintains contact with the road can be calculated from the equation \(v = \sqrt{gR}\). Remember to use consistent units throughout these calculations.

Step by step solution

01

Construct the Equation of Force

The downward force that the car exerts on the road is the difference between its weight and the centripetal force. Therefore, the equation of measurement becomes \(F = mg - mv^2/R \).
02

Create a Graph and Find the Line of Best Fit

Graph both sides of the equation with the speed \(v\)^2 on the x-axis and the force F on the y-axis. The graph will form a straight line with a slope equal to the value of \(m/R\) and y-intercept equal to \(mg\).
03

Calculate the Mass \(m\) and the Radius \(R\)

The slope of the graph is \(m/R\) and the y-intercept is \(mg\). Therefore, the mass of the car can be calculated by dividing the y-intercept by \(g\), the acceleration due to gravity. Then, calculate the radius of the arc \(R\) which will equal to \(m\) divided by the slope.
04

Calculate the Maximum Speed at which the Car will Lose Contact with the Road

The car will lose contact when the gravitational force equals the centripetal force, or in other words, when \(F = 0\). Solving the equation \(F = mg - mv^2/R \) for \(v\) when \(F = 0\), we get the maximum speed at which the car will lose contact with the road.

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Key Concepts

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

Circular Motion
When thinking about circular motion, imagine an object moving along a circular path. In our problem, the car travels over a hill, resembling part of a circular arc. This type of motion is key to understanding forces acting on the car as it goes over the hill.

Circular motion involves both centripetal force and velocity. The centripetal force keeps the object moving in a circle, always pointing inward, towards the center of the circle. It's important to remember that without this force, the object would not follow a circular path.

The velocity, on the other hand, is always tangent to the path, meaning it points along the direction of motion. When the car is on top of the hill, gravity acts as a centripetal force to keep it on track. A delicate balance occurs as the downward gravitational force and the centripetal force maintain the car's circular movement over the hill.

  • **Centripetal Force:** Essential for maintaining circular motion, always directed inward.
  • **Velocity:** Tangential to the path, keeps the object moving.
  • **Gravity as a Factor:** Acts as a part of the centripetal force in this scenario.
Kinematics
Kinematics helps us understand how objects move, providing us with a mathematical relationship between displacement, speed, velocity, and acceleration. In circular motion, as in this exercise, we're interested in how speed transforms motion through its impact on forces.

When plotting the speed squared against force, as instructed in the exercise, the graph reveals a linear relationship. This reflects the equation: \[ F = mg - \frac{mv^2}{R} \]

As speed increases, the centripetal force required also increases, changing how much force the car exerts downward on the road. This understanding helps us solve for unknown variables like the car's mass \(m\) and the radius \(R\) of the hill's arc. The change in the plotted values also guides us to determine at what speed the car will lose contact with the surface due to insufficient gravitational force.

  • **Understanding Motion:** Clarified through relationships of speed, gravity, and force.
  • **Graphical Representation:** Transform equations into visual data to find patterns.
  • **Acceleration and Force:** Key areas unfolding how kinematics explains movement.
Newton's Laws
Newton's Laws of Motion provide the backbone for understanding the dynamics of circular motion in this scenario.

**Newton's First Law** teaches us that an object in motion remains in motion unless acted upon by a net external force. In the car's scenario, it's constantly acted on by gravity and the ground's normal force because of the hill.

**Newton's Second Law** states that the force needed to keep an object in circular motion is provided by the centripetal force, which equals mass times acceleration \( F = ma \). In this case, the acceleration is the centripetal acceleration \( a = \frac{v^2}{R} \). Combining this with gravitational forces results in the equation used in the exercise to explore the balance of forces at the hill's crest.

**Newton's Third Law** reminds us that forces are interactions. As the car pushes on the road, the road offers an equal and opposite reaction force. This interaction is vital when plotting the force exerted against speed to grasp the relationships described in the exercise.

  • **First Law:** Stability in motion until external force intervenes.
  • **Second Law:** Relation of force, mass, and centripetal acceleration.
  • **Third Law:** Equal and opposite forces impact motion dynamics.

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Most popular questions from this chapter

Force on a Skater's Wrist. A \(52 \mathrm{~kg}\) ice skater spins about a vertical axis through her body with her arms horizontally outstretched; she makes 2.0 turns each second. The distance from one hand to the other is \(1.50 \mathrm{~m} .\) Biometric measurements indicate that each hand typically makes up about \(1.25 \%\) of body weight. (a) Draw a free-body diagram of one of the skater's hands. (b) What horizontal force must her wrist exert on her hand? (c) Express the force in part (b) as a multiple of the weight of her hand.

A \(75.0 \mathrm{~kg}\) wrecking ball hangs from a uniform, heavy-duty chain of mass \(26.0 \mathrm{~kg}\). (a) Find the maximum and minimum tensions in the chain. (b) What is the tension at a point three-fourths of the way up from the bottom of the chain?

DATA You are an engineer working for a manufacturing company. You are designing a mechanism that uses a cable to drag heavy metal blocks a distance of \(8.00 \mathrm{~m}\) along a ramp that is sloped at \(40.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between these blocks and the incline is \(\mu_{\mathrm{k}}=0.350 .\) Each block has a mass of \(2170 \mathrm{~kg}\). The block will be placed on the bottom of the ramp, the cable will be attached, and the block will then be given just enough of a momentary push to overcome static friction. The block is then to accelerate at a constant rate to move the \(8.00 \mathrm{~m}\) in \(4.20 \mathrm{~s}\). The cable is made of wire rope and is parallel to the ramp surface. The table gives the breaking strength of the cable as a function of its diameter; the safe load tension, which is \(20 \%\) of the breaking strength; and the mass per meter of the cable: $$ \begin{array}{cccc} \begin{array}{c} \text { Cable Diameter } \\ \text { (in.) } \end{array} & \begin{array}{c} \text { Breaking Strength } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Safe Load } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Mass per Meter } \\ (\mathbf{k g} / \mathbf{m}) \end{array} \\ \hline \frac{1}{4} & 24.4 & 4.89 & 0.16 \\ \frac{3}{8} & 54.3 & 10.9 & 0.36 \\ \frac{1}{2} & 95.2 & 19.0 & 0.63 \\ \frac{5}{8} & 149 & 29.7 & 0.98 \\ \frac{3}{4} & 212 & 42.3 & 1.41 \\ \frac{7}{8} & 286 & 57.4 & 1.92 \\ 1 & 372 & 74.3 & 2.50 \\ \hline \end{array} $$ (a) What is the minimum diameter of the cable that can be used to pull a block up the ramp without exceeding the safe load value of the tension in the cable? Ignore the mass of the cable, and select the diameter from those listed in the table. (b) You need to know safe load values for diameters that aren't in the table, so you hypothesize that the breaking strength and safe load limit are proportional to the cross-sectional area of the cable. Draw a graph that tests this hypothesis, and discuss its accuracy. What is your estimate of the safe load value for a cable with diameter \(\frac{9}{16}\) in.? (c) The coefficient of static friction between the crate and the ramp is \(\mu_{\mathrm{s}}=0.620,\) which is nearly twice the value of the coefficient of kinetic friction. If the machinery jams and the block stops in the middle of the ramp, what is the tension in the cable? Is it larger or smaller than the value when the block is moving? (d) Is the actual tension in the cable, at its upper end, larger or smaller than the value calculated when you ignore the mass of the cable? If the cable is \(9.00 \mathrm{~m}\) long, how accurate is it to ignore the cable's mass?

Apparent Weight. A \(550 \mathrm{~N}\) physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is \(850 \mathrm{~kg}\). As the elevator starts moving, the scale reads \(450 \mathrm{~N}\). (a) Find the acceleration of the elevator (magnitude and direction). (b) What is the acceleration if the scale reads \(670 \mathrm{~N}\) ? (c) If the scale reads zero, should the student worry? Explain. (d) What is the tension in the cable in parts (a) and (c)?

You are sitting on the edge of a horizontal disk (for example, a playground merry-go-round) that has radius \(3.00 \mathrm{~m}\) and is rotating at a constant rate about a vertical axis. (a) If the coefficient of static friction between you and the surface of the disk is \(0.400,\) what is the minimum time for one revolution of the disk if you are not to slide off? (b) Your friend's weight is half yours. If the coefficient of static friction for him is the same as for you, what is the minimum time for one revolution if he is not to slide off?

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