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Chapter 6: Problem 32

A uniform rectangular marble slab is \(3.4 \mathrm{~m}\) long and \(2.0 \mathrm{~m}\) wide. It has a mass of \(180 \mathrm{~kg}\). It is originally lying on the flat ground with its \(3.4-\mathrm{m} \times 2.0\) -m surface facing up. How much work is needed to stand it on its short end? [Hint: Think about its center of gravity.]

Short Answer

Expert verified
The work needed is 1764 J.

Step by step solution

01

Understand the Problem

To solve the problem, we need to calculate the work required to lift the center of gravity of the slab from its current position (when it is lying flat) to the new position (when it is standing on its short end).
02

Determine the Initial and Final Positions of the Center of Gravity

Initially, the center of gravity of the slab is at half the height of its thickness above the ground (since it's uniformly rectangular), which isn't given, so we assume it's negligible for the calculation. When the slab stands on its short end (2.0 m wide), the center of gravity will be at half the length of the current width above the ground, which is 1.0 m.
03

Calculate the Height Difference for the Center of Gravity

Since the center of gravity starts from near the ground (negligible height) and moves to 1.0 m above, the height difference is 1.0 m.
04

Use the Work-Energy Principle

Work done to lift the center of gravity = Mass \( \times \) Gravitational Acceleration \( \times \) Height Difference. Here, \( m = 180 \, \text{kg} \) and \( g \approx 9.8 \, \text{m/s}^2 \).
05

Calculate the Work Done

\[\text{Work Done} = 180 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1.0 \, \text{m} = 1764 \, \text{J}.\]

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a cornerstone of physics that connects the concepts of work and energy. It states that the work done on an object is equal to the change in its kinetic energy. However, in this particular exercise, we focus on potential energy since we're lifting a marble slab.

The formula for work done when lifting an object is given by:
  • Work Done = Force \( \times \) Distance
  • In terms of lifting, this becomes Work Done = Mass \( \times \) Gravitational force \( \times \) Height difference
This illustrates how much energy is needed to move an object vertically against the force of gravity. In our solution, we calculate the work needed to stand the slab by lifting its center of gravity from its original lying position to the standing position. The change in potential energy, which equates to the work done, is impacted by gravitational strength and height moved.
Understanding this principle helps you see how forces translate into energy changes, vital in many real-world scenarios.
Uniform Rectangular Slab
A uniform rectangular slab refers to a solid object with a consistent shape and density throughout. This slab is identified by its uniformity, meaning its mass is evenly distributed across its entire volume. For a uniform rectangular slab:
  • Shape: Determined by its dimensions (length, width, and thickness)
  • Center of Gravity: Located at the geometric center due to uniform mass distribution
  • Mass Distribution: This impacts how the slab behaves when moved or rotated
The slab in the exercise has specific dimensions: \(3.4\ \, \mathrm{m}\) long and \(2.0\ \, \mathrm{m}\) wide. Understanding the uniform nature helps in predicting how it will behave when force is applied to change its position. For example, when rotating the slab to stand on its short end, only the position of the center of gravity changes significantly due to its uniform mass distribution.
Gravitational Acceleration
Gravitational acceleration, typically denoted by \( g\), is a measure of how strongly gravity pulls objects towards Earth鈥檚 surface. The standard value used in calculations is \( 9.8 \, \mathrm{m/s^2}\), but this can vary slightly depending on location. It's crucial when calculating forces involved in lifting or moving objects against gravity.

In the exercise scenario, gravitational acceleration plays a key role in determining the work done to raise the slab. The force needed to lift, or in this case, tilt the slab upwards is directly influenced by gravitational pull, calculated as:
  • Gravitational Force = Mass \( \times \) Gravitational Acceleration
  • Height Difference: 1.0 m (as calculated in the solution)
Using these factors, we calculate work considering the slab鈥檚 weight and how high its center of gravity is lifted. This measurement helps us understand the energy expenditure in lifting tasks and informs any effort against gravity in engineering and daily applications.

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Most popular questions from this chapter

Compute the work done against gravity by a pump that discharges 600 liters of fuel oil into a tank \(20 \mathrm{~m}\) above the pump's intake. One cubic centimeter of fuel oil has a mass of \(0.82 \mathrm{~g} .\) One liter is \(1000 \mathrm{~cm}^{3}\) The mass lifted is $$ (600 \text { liters })\left(1000 \frac{\mathrm{cm}^{3}}{\text { liter }}\right)\left(0.82 \frac{\mathrm{g}}{\mathrm{cm}^{3}}\right)=492000 \mathrm{~g}=492 \mathrm{~kg} $$ The lifting work is then Work \(=(m g)(h)=\left(492 \mathrm{~kg} \times 9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20 \mathrm{~m})=96 \mathrm{~kJ}\)

A 10.0-kg block is launched up a \(30.0^{\circ}\) inclined plane at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). As it slides it loses \(200 \mathrm{~J}\) to friction. How far along the incline will it travel before coming to rest?

An advertisement claims that a certain 1200 -kg car can accelerate from rest to a speed of \(25 \mathrm{~m} / \mathrm{s}\) in a time of \(8.0 \mathrm{~s}\). What average power must the motor develop to produce this acceleration? Give your answer in both watts and horsepower. Ignore friction losses. The work done in accelerating the car is $$ \text { Work done }=\text { Change in } \mathrm{KE}=\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)=\frac{1}{2} m v_{f}^{2} $$ The time taken for this work to be performed is \(8.0 \mathrm{~s}\). Therefore, to two significant figures, $$ \text { Power }=\frac{\text { Work }}{\text { Time }}=\frac{\frac{1}{2}(1200 \mathrm{~kg})(25 \mathrm{~m} / \mathrm{s})^{2}}{8.0 \mathrm{~s}}=46875 \mathrm{~W}=47 \mathrm{~kW} $$ Converting from watts to horsepower, we have $$ \text { Power }=(46875 \mathrm{~W})\left(\frac{1 \mathrm{hp}}{746 \mathrm{~W}}\right)=63 \mathrm{hp} $$

A 60 000-kg train is being dragged along a straight line up a \(1.0\) percent grade (i.e., the road rises \(1.0 \mathrm{~m}\) for each \(100 \mathrm{~m}\) traveled horizontally) by a steady drawbar pull of \(3.0 \mathrm{kN}\) parallel to the incline. The friction force opposing the motion of the train is \(4.0\) \(\mathrm{kN}\). The train's initial speed is \(12 \mathrm{~m} / \mathrm{s}\). Through what distance s will the train move along its tracks before its speed is reduced to \(9.0 \mathrm{~m} / \mathrm{s}\) ? Use energy considerations. The change in total energy of the train is due to the work done by the friction force (which is negative) and the drawbar pull (which is positive): Change in \(\mathrm{KE}+\) change in \(\mathrm{PE}_{\mathrm{G}}=W_{\text {drawbar }}+W_{\text {friction }}\) The train loses \(\mathrm{KE}\) and gains \(\mathrm{PE}_{\mathrm{G}}\). It rises a height \(h=s \sin \theta\), where \(\theta\) is the incline angle and \(\tan \theta=1 / 100\). Hence, \(\theta=0.573^{\circ}\), and \(h=0.010 \mathrm{~s}\) (at small angles \(\tan \theta \approx \sin \theta\) ). Therefore, $$ \begin{array}{c} \frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)+m g(0.010 s)=(3000 \mathrm{~N})(s)(1)+(4000 \mathrm{~N})(s)(-1) \\ -1.89 \times 10^{6} \mathrm{~J}+\left(5.89 \times 10^{3} \mathrm{~N}\right) s=(-1000 \mathrm{~N}) s \end{array} $$ from which we obtain \(s=274 \mathrm{~m}=0.27 \mathrm{~km}\).

A narrowing ladder \(3.0 \mathrm{~m}\) long weighing \(200 \mathrm{~N}\) has its center of gravity \(120 \mathrm{~cm}\) from the bottom. At its top end is a \(50-\mathrm{N}\) can of paint. Compute the work required to raise the ladder from being horizontal, lying on the ground, to being vertical with its legs resting on the ground. In other words, how much work must be done to rotate the ladder into an upright vertical position, thereby raising both its center of gravity and the can of paint? The work done (against gravity) consists of two parts: the work to raise the center of gravity \(1.20 \mathrm{~m}\) and the work to raise the load at the end through \(3.0 \mathrm{~m}\). Therefore, Work done \(=(200 \mathrm{~N})(1.20 \mathrm{~m})+(50 \mathrm{~N})(3.0 \mathrm{~m})=0.39 \mathrm{~kJ}\)
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