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Chapter 6: Problem 30

How much work in total must a 200-lb man do climbing to the top of the 555 -ft-tall Washington Monument carrying a \(10.0\) -kg backpack? [Hint: \(1 \mathrm{lb}=4.448 \mathrm{~N}\).]

Short Answer

Expert verified
The total work done is 167,145.851 Joules.

Step by step solution

01

Convert Weight to Newtons

First, convert the man's weight from pounds to Newtons using the given conversion factor. The man weighs 200 lb, and 1 lb is equivalent to 4.448 N. Therefore, the force due to the man's weight is:\[ 200 ext{ lb} \times 4.448 \frac{\text{N}}{\text{lb}} = 889.6 \text{ N} \]
02

Calculate Force of Backpack in Newtons

Next, calculate the force exerted by the backpack's weight in Newtons.The mass of the backpack is 10.0 kg, and the gravitational force can be calculated using the formula: \[ W = mg \]where \( m = 10.0 \) kg and \( g = 9.81 \frac{\text{m}}{\text{s}^2} \).\[ W = 10.0 \text{ kg} \times 9.81 \frac{\text{m}}{\text{s}^2} = 98.1 \text{ N} \]
03

Sum the Total Force

Add together the force from the man and the force from the backpack to get the total force that needs to be exerted:\[ 889.6 \text{ N} + 98.1 \text{ N} = 987.7 \text{ N} \]
04

Calculate Work Done

Now, calculate the total work done against gravity. Work (W) is calculated as:\[ W = Fd \]where \( F \) is the total force (987.7 N) and \( d \) is the height of the climb (555 ft). First, convert the height to meters (as 1 ft = 0.3048 m):\[ 555 \text{ ft} \times 0.3048 \frac{\text{m}}{\text{ft}} = 169.164 \text{ m} \]Finally, calculate the work:\[ W = 987.7 \text{ N} \times 169.164 \text{ m} = 167,145.851 \text{ J} \]
05

Conclude the Calculation

The total work done by the man carrying the backpack to the top of the monument is approximately 167,145.851 Joules.

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Key Concepts

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

Force Conversion
When dealing with different units of measurement, such as pounds and Newtons, it's necessary to convert them so they can be used correctly in calculations. Here, we're handling both the weight of the man and a scientific concept like gravitational force.
To convert from pounds to Newtons, use the conversion factor: 1 pound equals 4.448 Newtons. Multiply the total weight in pounds by 4.448 to get the force in Newtons. For instance, a 200 lb man translates to:
  • 200 lb x 4.448 N/lb = 889.6 N
This conversion allows us to use consistent units when calculating force, which simplifies the process and maintains accuracy.
Weight Calculation
Weight is actually the force gravity exerts on an object, calculated by the formula:
  • \( W = mg \)
With this formula, you'll multiply the mass (\( m \), in kilograms) by the acceleration due to gravity (\( g \), which is approximately 9.81 m/s虏 on Earth).
For example, a backpack quite commonly weighs 10 kg. Hence, its weight is calculated as:
  • 10 kg x 9.81 m/s虏 = 98.1 N
Knowing how to calculate weight ensures precise evaluations when dealing with forces and movements, like lifting or carrying objects upward.
Work Calculation
The concept of work in physics refers to how much energy is transferred when a force is applied over a distance. Calculating work involves the formula:
  • \( W = Fd \)
where \( F \) stands for force (measured in Newtons), and \( d \) is the distance over which the force acts (measured in meters). In this case, the Washington Monument climb height needs converting from feet to meters.
  • 555 ft = 169.164 m (using 1 ft = 0.3048 m)
Now, calculate work by multiplying the total force (987.7 N) by the calculated distance (169.164 m):
  • Work = 987.7 N x 169.164 m = 167,145.851 J
Thus, the effort to climb the monument while bearing additional weight is quantified in Joules, providing a clear picture of energy expenditure.
Gravitational Force
Gravitational force stands as a fundamental interaction that compels masses to attract one another. On Earth, this manifests predominantly as the force pulling objects towards the ground. The acceleration due to gravity (\( g \)) is about 9.81 m/s虏.
Here鈥檚 how gravitational force factors into weight calculations:
  • The force exerted by an object equals its weight, computed as \( mg \).
  • For a 10 kg backpack, this results in a force of 98.1 N, as discussed before.
Administering a broader scope, gravitational force is essential to energy calculations and impacts how we measure and predict the work needed to change an object's vertical position.

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Most popular questions from this chapter

A \(1200-\mathrm{kg}\) car coasts from rest down a driveway that is inclined \(20^{\circ}\) to the horizontal and is \(15 \mathrm{~m}\) long. How fast is the car going at the end of the driveway if \((a)\) friction is negligible and \((b)\) a friction force of \(3000 \mathrm{~N}\) opposes the motion?

At sea level a nitrogen molecule in the air has an average translational KE of \(6.2 \times 10^{-21} \mathrm{~J}\). Its mass is \(4.7 \times 10^{-26} \mathrm{~kg} .(a)\) If such a molecule could shoot straight up without striking other air molecules, how high would it rise? ( \(b\) ) What is that molecule's initial upward speed?

An engine expends \(40.0\) hp in propelling a car along a level track at a constant speed of \(15.0 \mathrm{~m} / \mathrm{s}\). How large is the total retarding force acting on the car? Remember that \(1 \mathrm{hp}=745.7 \mathrm{~W}\).

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 \(s=274 \mathrm{~m}=0.27 \mathrm{~km}\).

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 \(=(m g)(h)=\left(492 \mathrm{~kg} \times 9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20 \mathrm{~m})=96 \mathrm{~kJ}\)
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