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Chapter 8: Problem 27

Tarzan, who weighs 688 N, swings from a cliff at the end of a vine \(18 \mathrm{~m}\) long \((\) Fig. \(8-40) .\) From the top of the cliff to the bottom of the swing, he descends by \(3.2 \mathrm{~m}\). The vine will break if the force on it exceeds \(950 \mathrm{~N}\). (a) Does the vine break? (b) If no, what is the greatest force on it during the swing? If yes, at what angle with the vertical does it break?

Short Answer

Expert verified
The vine does not break; the greatest force is less than 950 N.

Step by step solution

01

Determine Potential Energy Change

First, calculate the change in Tarzan's gravitational potential energy as he descends the height of 3.2 m. Since potential energy is given by the formula \( PE = mgh \), where \( m \) is the mass (and given as weight \( 688 \, N \)), \( g = 9.8 \, m/s^2 \), and \( h = 3.2 \, m \), the potential energy change is \( \Delta PE = mgh = 688 \times 3.2 \).
02

Calculate Kinetic Energy at the Bottom

The kinetic energy Tarzan has at the bottom of the swing is the energy converted from potential energy, so \( KE = \Delta PE = 688 \times 3.2 \) Joules.
03

Apply Conservation of Energy

Using conservation of mechanical energy, \( \frac{1}{2}mv^2 = mgh \), rearrange and solve for \( v \), the velocity at the bottom of the swing: \( \frac{1}{2} v^2 = gh \rightarrow v^2 = 2gh \rightarrow v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 3.2} \).
04

Calculate Tension at the Bottom of the Swing

The tension in the vine at the bottom of the swing is the sum of the gravitational force and the centripetal force required to maintain Tarzan's circular path. Centripetal force \( F_c \) is given by \( F_c = \frac{mv^2}{r} \), where \( r = 18 \, m \). Therefore, the total tension \( T \) is \( T = mg + \frac{mv^2}{r} \). Substitute the values to find \( T = 688 + \frac{688 \cdot v^2}{18 \times 9.8} \).
05

Check If Vine Breaks

Compare the calculated tension at the bottom of the swing to the breaking force of 950 N. If \( T \leq 950 \), the vine does not break. Calculate \( T \) to determine the outcome.

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Key Concepts

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

Potential Energy
Potential energy, often abbreviated as PE, is a fundamental concept in understanding how energy is stored within an object due to its position relative to other objects. In the case of Tarzan swinging from a vine, his potential energy is highest when he is at the starting point on the cliff. This is because potential energy depends on height and is calculated using the formula:\[PE = mgh\]where:
  • \(m\) is the mass of the object,
  • \(g\) is the acceleration due to gravity (9.8 m/s虏 on Earth),
  • \(h\) is the height above the reference point, which in this exercise is 3.2 meters below the top of the cliff.
As Tarzan swings down, his potential energy decreases because he is losing height. This loss in potential energy is converted into kinetic energy, which propels him forward. Understanding potential energy helps in predicting how much kinetic energy he can gain as he swings lower.
Kinetic Energy
Kinetic energy refers to the energy an object possesses due to its motion. The faster an object moves, the more kinetic energy it has. For Tarzan, as he descends from the cliff, the potential energy lost due to decrease in height becomes his kinetic energy.The formula to calculate kinetic energy (KE) is\[KE = \frac{1}{2}mv^2\]Since Tarzan's potential energy was fully transformed into kinetic energy at the bottom of his swing, it means:\[KE = \Delta PE\]This equation marks the energy conservation principle at work. By solving for Tarzan's velocity using the formula \( v = \sqrt{2gh}\), it helps us understand how energy transformations work in dynamic systems, leading to the calculation of velocity and further insights into forces acting upon him during his swing.
Centripetal Force
Centripetal force plays a crucial role when an object, like Tarzan, moves along a curved path. It is the force that keeps the object moving in a circle and is directed towards the center of the circular path. For Tarzan, as he swings on the vine, this force is necessary to maintain his circular arc.The centripetal force \(F_c\) is described by the formula:\[F_c = \frac{mv^2}{r}\]where:
  • \(m\) is the mass of the object,
  • \(v\) is the velocity of the object,
  • \(r\) is the radius of the circular path, which is the vine's length, 18 meters.
This force, in combination with gravitational force, determines the tension in the vine at the bottom of Tarzan's swing. Calculating the total tension involves adding the gravitational force and the centripetal force, ensuring that Tarzan stays on his path without the vine breaking. Evaluating whether the vine exceeds its strength is crucial to understanding the limits of materials and forces in mechanical systems.

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Most popular questions from this chapter

The magnitude of the gravitational force between a particle of mass \(m_{1}\) and one of mass \(m_{2}\) is given by $$ F(x)=G \frac{m_{1} m_{2}}{x^{2}} $$ where \(G\) is a constant and \(x\) is the distance between the particles. (a) What is the corresponding potential energy function \(U(x) ?\) Assume that \(U(x) \rightarrow 0\) as \(x \rightarrow \infty\) and that \(x\) is positive. (b) How much work is required to increase the separation of the particles from \(x=x_{1}\) to \(x=x_{1}+d ?\)

A block with mass \(m=2.00 \mathrm{~kg}\) is placed against a spring on a frictionless incline with angle \(\theta=30.0^{\circ}\) (Fig. 8-44). (The block is not attached to the spring.) The spring, with spring constant \(k=19.6 \mathrm{~N} / \mathrm{cm},\) is compressed \(20.0 \mathrm{~cm}\) and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block- Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

A skier weighing \(600 \mathrm{~N}\) goes over a frictionless circular hill of radius \(R=20 \mathrm{~m}\) (Fig. 8 -62). Assume that the effects of air resistance on the skier are negligible. As she comes up the hill, her speed is \(8.0 \mathrm{~m} / \mathrm{s}\) at point \(B,\) at angle \(\theta=20^{\circ} .\) (a) What is her speed at the hilltop (point \(A\) ) if she coasts without using her poles? (b) What minimum speed can she have at \(B\) and still coast to the hilltop? (c) Do the answers to these two questions increase, decrease, or remain the same if the skier weighs \(700 \mathrm{~N}\) instead of \(600 \mathrm{~N} ?\)

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant \(k=400 \mathrm{~N} / \mathrm{m} ;\) the other end of the spring is fixed in place. The cookie has a kinetic energy of \(20.0 \mathrm{~J}\) as it passes through the spring's equilibrium position. As the cookie slides, a frictional force of magnitude \(10.0 \mathrm{~N}\) acts on it. (a) How far will the cookie slide from the equilibrium position before coming momentarily to rest? (b) What will be the kinetic energy of the cookie as it slides back through the equilibrium position?

In a circus act, a \(60 \mathrm{~kg}\) clown is shot from a cannon with an initial velocity of \(16 \mathrm{~m} / \mathrm{s}\) at some unknown angle above the horizontal. A short time later the clown lands in a net that is \(3.9 \mathrm{~m}\) vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?
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