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Chapter 7: Problem 11

You are testing a new amusement park roller coaster with an empty car of mass 120 kg. One part of the track is a vertical loop with radius 12.0 m. At the bottom of the loop (point A) the car has speed 25.0 m>s, and at the top of the loop (point B) it has speed 8.0 m>s. As the car rolls from point A to point B, how much work is done by friction?

Short Answer

Expert verified
The work done by friction as the car rolls from point A to point B is 5445.6 Joules.

Step by step solution

01

Calculate the Kinetic Energy at Points A and B

First, use the kinetic energy formula \( KE = 0.5 * m * v^2 \) , where m is mass and v is velocity. At point A, calculate \( KE_A = 0.5 * 120 * (25.0)^2 = 37500 J \). At point B, do the same to get \( KE_B = 0.5 * 120 * (8.0)^2 = 3840 J \).
02

Determine the Gravitational Potential Energy at Points A and B

Use the formula for gravitational potential energy: \( PE = m * g * h \), where m is mass, g is gravitational acceleration (9.81 m/s^2 on the surface of the Earth), and h is height. At point A, the height is 0, so \( PE_A = 0 \). At point B (top of the loop), the height is twice the radius of the loop, so \( PE_B = 120 * 9.81 * 2 * 12 = 28214.4 J \).
03

Calculate the Total Energy at Points A and B

Add the kinetic and potential energy at both points using the formula \( TE = KE + PE \). At point A, \( TE_A = 37500 + 0 = 37500 J \). At point B, \( TE_B = 3840 + 28214.4 = 32054.4 J \).
04

Calculate the Work Done by Friction

The work done by friction is the difference in total energies between points A and B. Using the work-energy theorem, \( W = 螖KE + 螖PE \), compute for the work done: \( W = TE_A - TE_B = 37500 - 32054.4 = 5445.6 J \). This indicates that 5445.6 joules of work are done by friction as the car rolls from point A to point B.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. This type of energy depends on two main factors: the mass of the object and its velocity. The greater the mass and the faster the velocity, the more kinetic energy an object will have.
Kinetic energy can be calculated using the formula:
  • \( KE = \frac{1}{2} m v^2 \)
where:
  • \( m \) is the mass of the object
  • \( v \) is its velocity

In the context of our roller coaster problem, we calculate kinetic energy at two points: at the bottom of the loop (point A) and at the top of the loop (point B). At point A, the car has a higher speed (25 m/s), resulting in high kinetic energy (37,500 J). At point B, the speed decreases to 8 m/s, greatly reducing the kinetic energy to 3,840 J.
Understanding how kinetic energy changes as speed changes helps explain how the roller coaster moves through different parts of the track.
Potential Energy
Potential energy, especially gravitational potential energy, is the energy stored in an object due to its position relative to a gravitational source. For objects near Earth's surface, this energy can be calculated with:
  • \( PE = m g h \)
where:
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity (9.81 m/s虏 on Earth)
  • \( h \) is the height above the reference point

In the roller coaster problem, at point A, the car is at the lowest point (height is zero), thus its potential energy is also zero. At point B, at the top of the loop, the car is at its highest point, 24 meters above point A, giving it a significant potential energy of 28,214.4 J.
This increase in potential energy as the car ascends demonstrates how energy is transferred from kinetic energy to potential energy as the car climbs higher in the loop.
Work-Energy Theorem
The work-energy theorem relates the work done on an object to its change in kinetic and potential energy. Essentially, it states that the work done by all external forces acting on an object results in a change in the object鈥檚 total mechanical energy.
This theorem is represented as:
  • \( W = 螖KE + 螖PE \)
  • \( W \) is the work done by external forces
  • \( 螖KE \) is the change in kinetic energy
  • \( 螖PE \) is the change in potential energy

In our scenario, as the roller coaster moves from point A to point B, some energy is lost due to friction, which acts as an external force. By calculating the difference between the total mechanical energies at points A and B (37,500 J and 32,054.4 J respectively), we find that 5,445.6 J of work is done by friction.
The work-energy theorem provides a way to account for this energy loss and shows how energy transformations and work affect the motion of an object, like the roller coaster car.

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Most popular questions from this chapter

A small rock with mass 0.20 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with radius R = 0.50 m (Fig. E7.9). Assume that the size of the rock is small com-pared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls. The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl has magnitude 0.22 J. (a) Between points A and B, how much work is done on the rock by (i) the normal force and (ii) gravity? (b) What is the speed of the rock as it reaches point B? (c) Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? Explain. (d) Just as the rock reaches point B, what is the normal force on it due to the bottom of the bowl?

Tarzan and Jane. Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length \(20 \mathrm{~m}\) that makes an angle of \(45^{\circ}\) with the vertical, steps off his tree limb, and swings down and then up to Jane's open arms. When he arrives, his vine makes an angle of \(30^{\circ}\) with the vertical. Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan's speed just before he reaches Jane. Ignore air resistance and the mass of the vine.

A block with mass \(0.50 \mathrm{~kg}\) is forced against a horizontal spring of negligible mass, compressing the spring a distance of \(0.20 \mathrm{~m}\) (Fig. \(\mathbf{P 7 . 3 9}\) ). When released, the block moves on a horizontal tabletop for \(1.00 \mathrm{~m}\) before coming to rest. The force constant \(k\) is \(100 \mathrm{~N} / \mathrm{m}\). What is the coefficient of kinetic friction \(\mu_{\mathrm{k}}\) between the block and the tabletop?

A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x)=\left(\alpha / x^{2}\right)-(\beta / x),\) where \(\alpha\) and \(\beta\) are positive constants. The proton is released from rest at \(x_{0}=\alpha / \beta\). (a) Show that \(U(x)\) can be written as $$ Graph \(U(x) .\) Calculate \(U\left(x_{0}\right)\) and thereby locate the point \(x_{0}\) on the graph. (b) Calculate \(v(x)\), the speed of the proton as a function of position. Graph \(v(x)\) and give a qualitative description of the motion. (c) For what value of \(x\) is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the proton be released instead at \(x_{1}=3 \alpha / \beta .\) Locate the point \(x_{1}\) on the graph of \(U(x) .\) Calculate \(v(x)\) and give a qualitative description of the motion. (f) For each release point \(\left(x=x_{0}\right.\) and \(\left.x=x_{1}\right),\) what are the maximum and minimum values of \(x\) reached during the motion? U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right] $$

A conservative force \(\vec{F}\) is in the \(+x\) -direction and has magnitude \(F(x)=\alpha /\left(x+x_{0}\right)^{2},\) where \(\alpha=0.800 \mathrm{~N} \cdot \mathrm{m}^{2}\) and \(x_{0}=0.200 \mathrm{~m}\). (a) What is the potential- energy function \(U(x)\) for this force? Let \(U(x) \rightarrow 0\) as \(x \rightarrow \infty .\) (b) An object with mass \(m=0.500 \mathrm{~kg}\) is released from rest at \(x=0\) and moves in the \(+x\) -direction. If \(\vec{F}\) is the only force acting on the object, what is the object's speed when it reaches \(x=0.400 \mathrm{~m} ?\)
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