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Chapter 7: Problem 37

A \(700-\mathrm{N}\) Marine in basic training climbs a \(10.0-\mathrm{m}\) vertical rope at a constant speed in 8.00 s. What is his power output?

Short Answer

Expert verified
The power output of the Marine can be calculated by substituting the calculated Work Done and given Time taken into the Power formula.

Step by step solution

01

Calculate the Work Done by the Marine

The work done by the Marine can be calculated using the formula for gravitational potential energy which is \( W = mgh \). Here, m is the weight of the Marine in Newtons converted to mass, g is acceleration due to gravity, and h is the height climbed. The weight of the Marine is given as 700 N and the height of the rope is given as 10 m. However, we will first need to convert the weight in Newtons to mass in kilograms by dividing by the acceleration due to gravity (g), which is approximately 9.8 m/s^2. So, the mass will be \( m = \frac{700 \, N}{ 9.8 \, m/s^2} \) and the work done will be given by \( W = mgh \).
02

Calculate the Power Output

Power is defined as the work done per unit time. Given that we have already calculated the work done and we know the time taken (8 seconds), we can find the power output with the formula \( P = \frac{W}{t} \).
03

Substitute the Values into the Power Formula and Find the Power Output

Substitute the calculated value for work and the given time into the power formula to find the power output.

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Key Concepts

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

Work Done
When we talk about work done in physics, it refers to the energy transferred when a force moves an object over a distance. In this case, the Marine climbing the rope exerts a force against gravity. This force moves the Marine vertically, thereby doing work. Work done (W) is calculated using the formula:
\[ W = Fd \]
Where:
  • \( F \) is the force applied, in this exercise equal to the Marine's weight, 700 N.
  • \( d \) is the distance moved in the direction of the force, which is 10 m.
Since the Marine climbs vertically, the work done equates to changes in gravitational potential energy, helping us understand energy exchanges during such efforts.
Gravitational Potential Energy
Gravitational Potential Energy (GPE) is the energy an object possesses due to its position in a gravitational field. In our scenario, as the Marine ascends the rope, he gains height and thus, his gravitational potential energy increases. GPE can be calculated using:
\[ GPE = mgh \]
Where:
  • \( m \) is the mass of the Marine, which we find by dividing his weight (700 N) by the gravitational acceleration (9.8 m/s虏).
  • \( g \) is the acceleration due to gravity \((9.8 m/s虏)\).
  • \( h \) is the height climbed \((10 m)\).
Gravitational potential energy plays a pivotal role in this exercise as it equates to the work done by the Marine while climbing, thus providing us with the necessary variables to calculate work done and eventually the power output.
Constant Speed
Constant speed means an object covers equal distances in equal time intervals, implying no net acceleration. For the Marine, climbing the rope at constant speed suggests that upward force exerted equals the weight, maintaining equilibrium. This characteristic simplifies calculations because:
  • It indicates that all the work done is effective in raising the Marine鈥檚 gravitational potential energy.
  • The absence of acceleration lets us directly calculate the work done and power based only on distance and force.
When calculating power output, the uniform speed allows us to confidently use the Marine's weight and rope height to measure energy transitions without considering additional forces or energy losses beyond gravitational resistance.

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Most popular questions from this chapter

A particle moves along the \(x\) axis from \(x=12.8 \mathrm{m}\) to \(x=23.7 \mathrm{m}\) under the influence of a force $$F=\frac{375}{x^{3}+3.75 x}$$ where \(F\) is in newtons and \(x\) is in meters. Using numerical integration, determine the total work done by this force on the particle during this displacement. Your result should be accurate to within \(2 \%.\)

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.

For \(\mathbf{A}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{B}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}},\) and \(\mathbf{C}=2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\) find \(\mathbf{C} \cdot(\mathbf{A}-\mathbf{B}).\)

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?

A bead at the bottom of a bowl is one example of an object in a stable equilibrium position. When a physical system is displaced by an amount \(x\) from stable equilibrium, a restoring force acts on it, tending to return the system to its equilibrium configuration. The magnitude of the restoring force can be a complicated function of \(x .\) For example, when an ion in a crystal is displaced from its lattice site, the restoring force may not be a simple function of \(x .\) In such cases we can generally imagine the function \(F(x)\) to be expressed as a power series in \(x,\) as \(F(x)=-\left(k_{1} x+k_{2} x^{2}+k_{3} x^{3}+\ldots\right) .\) The first term here is just Hooke's law, which describes the force exerted by a simple spring for small displacements. For small excursions from equilibrium we generally neglect the higher order terms, but in some cases it may be desirable to keep the second term as well. If we model the restoring force as \(F=-\left(k_{1} x+k_{2} x^{2}\right),\) how much work is done in displacing the system from \(x=0\) to \(x=x_{\max }\) by an applied force \(-F ?\)
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