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Chapter 7: Problem 3

Batman, whose mass is \(80.0 \mathrm{kg},\) is dangling on the free end of a \(12.0-\mathrm{m}\) rope, the other end of which is fixed to a tree limb above. He is able to get the rope in motion as only Batman knows how, eventually getting it to swing enough that he can reach a ledge when the rope makes a \(60.0^{\circ}\) angle with the vertical. How much work was done by the gravitational force on Batman in this maneuver?

Short Answer

Expert verified
The work done by the gravitational force on Batman in this maneuver is \( 4704.0 \, \mathrm{J} \).

Step by step solution

01

Identify known and unknown values

The known values are Batman's mass \( m = 80.0 \, \mathrm{kg} \), the length of the rope \( L = 12.0 \, \mathrm{m} \) and the rope's maximum angle from the vertical \( \theta = 60.0^{\circ} \). The unknown value is the work done by the gravitational force \( W \).
02

Calculate the vertical displacement

From geometrical considerations, when the swing reaches its maximum angle, the vertical component of the displacement is \( \Delta y = L - L \cos \theta \). Substitute \( L = 12.0 \mathrm{m} \) and \( \theta = 60.0^{\circ} \) into the equation to calculate \( \Delta y \). The cosine of \( 60.0^{\circ} \) is \( 0.5 \). Thus, \( \Delta y = 12.0 \, \mathrm{m} - 12.0 \, \mathrm{m} \cdot 0.5 = 6.0 \, \mathrm{m} \).
03

Calculate the work done by the gravitational force

The work done by gravity is given by the formula \( W = m \cdot g \cdot \Delta y \), where \( g \) is the acceleration due to gravity (\( g = 9.8 \, \mathrm{m/s}^2 \)). Substitute \( m = 80.0 \, \mathrm{kg}, g = 9.8 \, \mathrm{m/s}^2 \), and \( \Delta y = 6.0 \, \mathrm{m} \) into the equation to calculate \( W \). Thus, \( W = 80.0 \, \mathrm{kg} \cdot 9.8 \, \mathrm{m/s}^2 \cdot 6.0 \, \mathrm{m} = 4704.0 \, \mathrm{J} \).

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Key Concepts

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

Gravitational Force
The gravitational force is a fundamental interaction in physics that attracts two bodies with mass. For everyday situations on Earth, this force is what we perceive as weight, and it's due to the Earth's gravitational pull. In the context of our Batman problem, the gravitational force exerted on Batman can be expressed by the equation:
\( F = m \cdot g \)
where \( m \) is the mass of the object (80.0 kg for Batman) and \( g \) is the acceleration due to gravity (9.8 m/s虏 on Earth). It's this force that is doing work on Batman as he swings on the rope. The concept of work in physics is tied to a force causing a displacement. Here, gravitational force does work as Batman is displaced vertically while swinging. Remember, work is a scalar quantity and involves both force and the distance over which it acts; hence, the trigonometry comes into play to find the vertical component of Batman's displacement.
Work-Energy Principle
The work-energy principle is a key concept in physics, relating the work done on an object to its kinetic and potential energy changes. The work done by the gravitational force, as in Batman's case, affects his potential energy. This principle can be expressed as:
\( W = \Delta KE + \Delta PE \)
where \( W \) is the work done on the object, \( \Delta KE \) is the change in kinetic energy, and \( \Delta PE \) is the change in potential energy. For Batman hanging still at the start and reaching the ledge without gaining or losing speed, the change in kinetic energy is zero, so the work done by gravity is entirely converted into potential energy. This leads us to calculate the work done as being equal to the change in potential energy, which is \( m \cdot g \cdot \Delta y \) in his scenario. Thus, determining the work done by gravitational force helps us understand how energy is transferred and transformed.
Trigonometry in Physics
Trigonometry, the branch of mathematics dealing with the relationships between the sides and angles of triangles, is essential in solving many physics problems. When Batman swings on a rope to reach a ledge, trigonometry helps us calculate his vertical displacement. This calculation involves understanding sine, cosine, and tangent functions of angles. In the Batman problem, the cosine function is used to find the vertical component of the displacement (\( \Delta y \)) as he swings to make a \(60.0^\circ\) angle with the vertical.
\( \Delta y = L - L \cos(\theta) \)
Knowing the initial and final positions of Batman and the angle of the swing allows us to map out the triangle formed and quantitatively relate the angle to the side lengths using trigonometry. This application underscores the significance of trigonometry in bridging the gap between abstract angles and practical displacement or distance measures in physics problems.

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Most popular questions from this chapter

A block of mass \(2.50 \mathrm{kg}\) is pushed \(2.20 \mathrm{m}\) along a frictionless horizontal table by a constant 16.0 -N force directed \(25.0^{\circ}\) below the horizontal. Determine the work done on the block by (a) the applied force, (b) the normal force exerted by the table, and (c) the gravitational force. (d) Determine the total work done on the block.

A shopper in a supermarket pushes a cart with a force of \(35.0 \mathrm{N}\) directed at an angle of \(25.0^{\circ}\) downward from the horizontal. Find the work done by the shopper on the cart as he moves down an aisle \(50.0 \mathrm{m}\) long.

A 100 -g bullet is fired from a rifle having a barrel \(0.600 \mathrm{m}\) long. Assuming the origin is placed where the bullet begins to move, the force (in newtons) exerted by the expanding gas on the bullet is \(15000+10000 x-25000 x^{2}\) where \(x\) is in meters. (a) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel. (b) What If? If the barrel is 1.00 m long, how much work is done, and how does this value compare to the work calculated in (a)?

Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as \(1 \mathrm{kcal}=4186 \mathrm{J} .\) Metabolizing one gram of fat can release 9.00 kcal. A student decides to try to lose weight by exercising. She plans to run up and down the stairs in a football stadium as fast as she can and as many times as necessary. Is this in itself a practical way to lose weight? To evaluate the program, suppose she runs up a flight of 80 steps, each \(0.150 \mathrm{m}\) high, in \(65.0 \mathrm{s}\). For simplicity, ignore the energy she uses in coming down (which is small). Assume that a typical efficiency for human muscles is \(20.0 \%\) This means that when your body converts \(100 \mathrm{J}\) from metabolizing fat, \(20 \mathrm{J}\) goes into doing mechanical work (here, climbing stairs). The remainder goes into extra internal energy. Assume the student's mass is \(50.0 \mathrm{kg} .\) (a) How many times must she run the flight of stairs to lose one pound of fat? (b) What is her average power output, in watts and in horsepower, as she is running up the stairs?

You can think of the work-kinetic energy theorem as a second theory of motion, parallel to Newton's laws in describing how outside influences affect the motion of an object. In this problem, solve parts (a) and (b) separately from parts \((\mathrm{c})\) and \((\mathrm{d})\) to compare the predictions of the two theories. In a rifle barrel, a \(15.0-\mathrm{g}\) bullet is accelerated from rest to a speed of \(780 \mathrm{m} / \mathrm{s}\). (a) Find the work that is done on the bullet. (b) If the rifle barrel is \(72.0 \mathrm{cm}\) long, find the magnitude of the average total force that acted on it, as \(F=W /(\Delta r \cos \theta) .\) (c) Find the constant acceleration of a bullet that starts from rest and gains a speed of \(780 \mathrm{m} / \mathrm{s}\) over a distance of \(72.0 \mathrm{cm} .\) (d) If the bullet has mass \(15.0 \mathrm{g},\) find the total force that acted on it as \(\Sigma F=m a.\)
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