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Chapter 6: Problem 85

A pump is required to lift 800 kg of water (about 210 gallons) per minute from a well 14.0 m deep and eject it with a speed of 18.0 m/s. (a) How much work is done per minute in lifting the water? (b) How much work is done in giving the water the kinetic energy it has when ejected? (c) What must be the power output of the pump?

Short Answer

Expert verified
(a) 109,872 J, (b) 129,600 J, (c) 3,991.2 W.

Step by step solution

01

Determine the Work Done in Lifting the Water

The work done to lift the water can be calculated using the formula for gravitational work:\[ W_1 = mgh \]where \( W_1 \) is the work done lifting the water, \( m \) is the mass of the water, \( g \) is the acceleration due to gravity (approximately \(9.81 \ \text{m/s}^2\)), and \( h \) is the height lifted.Substitute the values:\[ W_1 = 800 \ \text{kg} \times 9.81 \ \text{m/s}^2 \times 14.0 \ \text{m} \]\[ W_1 = 109,872 \ \text{J} \] So, the work done per minute in lifting the water is \(109,872 \ \text{J}\).
02

Calculate the Work Done in Giving the Water Kinetic Energy

The work done to give the water kinetic energy can be calculated using the kinetic energy formula:\[ W_2 = \frac{1}{2} mv^2 \]where \( W_2 \) is the work done, \( m \) is the mass of the water, and \( v \) is the velocity.Substitute the values:\[ W_2 = \frac{1}{2} \times 800 \ \text{kg} \times (18.0 \ \text{m/s})^2 \]\[ W_2 = \frac{1}{2} \times 800 \ \times 324 \]\[ W_2 = 129,600 \ \text{J} \]Hence, the work done in giving the water the kinetic energy is \(129,600 \ \text{J}\).
03

Calculate the Total Work Done

The total work done by the pump per minute is the sum of the work done to lift the water and the work done to give it kinetic energy:\[ W_{\text{total}} = W_1 + W_2 \]\[ W_{\text{total}} = 109,872 \ \text{J} + 129,600 \ \text{J} \]\[ W_{\text{total}} = 239,472 \ \text{J} \] Thus, the total work done is \(239,472 \ \text{J}\).
04

Determine the Power Output of the Pump

Power output of the pump is calculated by dividing the total work done by the time duration, which is 1 minute or 60 seconds:\[ P = \frac{W_{\text{total}}}{t} \]\[ P = \frac{239,472 \ \text{J}}{60 \ \text{s}} \]\[ P = 3,991.2 \ \text{W} \]Thus, the power output of the pump must be \(3,991.2 \ \text{W}\).

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Key Concepts

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

Gravitational Work
When it comes to lifting objects against gravity, the concept of gravitational work is crucial. In simple terms, gravitational work refers to the energy required to lift a mass to a certain height against the gravitational force. The formula to calculate this work is given by: \[ W = mgh \]
Here, \( W \) represents the work done, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth), and \( h \) is the height to which the object is lifted.
  • Mass: The amount of matter in the object.
  • Gravity: A constant that represents the force pulling objects toward the Earth's center.
  • Height: The vertical distance the object is moved against gravity.

In the original problem, lifting 800 kg of water to a height of 14 meters requires a certain amount of energy. Using the formula, you can calculate this as approximately 109,872 Joules. This energy expenditure is what keeps the water elevated and ready for its next phase of movement.
Kinetic Energy
Once the water is lifted, it must be ejected with velocity. Here is where kinetic energy comes into play. Kinetic energy is the energy that an object possesses due to its motion. The formula for kinetic energy is: \[ KE = \frac{1}{2} mv^2 \]
Where \( KE \) is the kinetic energy, \( m \) is the mass of the object, and \( v \) is the velocity at which the object is moving.
  • Mass: The weight of the water being ejected.
  • Velocity: The speed at which the water exits the pump.

For the given problem, water is ejected at a speed of 18 m/s. Hence, the kinetic energy given to the water can be calculated to be about 129,600 Joules. This energy is what propels the water as it exits, allowing it to move freely away from the pump.
Power Output
Power is defined as the rate at which work is done or energy is transferred over time. It helps us understand how quickly energy is used or generated. The formula to determine power is: \[ P = \frac{W}{t} \]
Where \( P \) is power, \( W \) is the total work done, and \( t \) is the time over which the work is done. In this case, the total work includes both the gravitational work and the kinetic energy.
  • Work: Total energy expended to lift and move the water.
  • Time: Duration of 60 seconds, equivalent to one minute.

By calculating the power output for the pump lifting and ejecting water, we find it to be approximately 3,991.2 Watts. Knowing the power output is essential for assessing how effectively and efficiently the pump operates over the set period.

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Most popular questions from this chapter

A 4.00-kg block of ice is placed against a horizontal spring that has force constant \(k\) = 200 N/m and is compressed 0.025 m. The spring is released and accelerates the block along a horizontal surface. Ignore friction and the mass of the spring. (a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. (b) What is the speed of the block after it leaves the spring?

A 5.00-kg package slides 2.80 m down a long ramp that is inclined at 24.0\(^\circ\) below the horizontal. The coefficient of kinetic friction between the package and the ramp is \(\mu_k\) \(=\) 0.310. Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity; (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.20 m/s at the top of the ramp, what is its speed after it has slid 2.80 m down the ramp?

We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M\), equilibrium length \(L_0\), and force constant \(k\). The work done to stretch or compress the spring by a distance \(L\) is \\(\frac{1}{2}\\) \(kX^2\), where \(X = L - L_0\). Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\). Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. (a) Calculate the kinetic energy of the spring in terms of \(M\) and \(v\). (\(Hint\): Divide the spring into pieces of length \(dl\); find the speed of each piece in terms of \(l\), \(v\), and \(L\); find the mass of each piece in terms of \(dl\), \(M\), and \(L\); and integrate from \(0\) to \(L\). The result is \(not\) \\(\frac{1}{2}\\) \(Mv^2\), since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 kg and force constant 3200 N/m is compressed 2.50 cm from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 0.053-kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

An object has several forces acting on it. One of these forces is \(\overrightarrow{F}= axy\hat{\imath}\), a force in the \(x\)-direction whose magnitude depends on the position of the object, with \(\alpha = 2.50 \, \mathrm{N/m}^2\). Calculate the work done on the object by this force for the following displacements of the object: (a) The object starts at the point (\(x = 0\), \(y = 3.00\) m) and moves parallel to the x-axis to the point (\(x= 2.00\) m, \(y = 3.00\) m). (b) The object starts at the point (\(x = 2.00\) m, \(y = 0\)) and moves in the \(y\)-direction to the point (\(x = 2.00\) m, \(y = 3.00\) m). (c) The object starts at the origin and moves on the line \(y = 1.5x\) to the point (\(x = 2.00\) m, \(y = 3.00\) m).

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do 80.0 J of work when you compress the springs 0.200 m from their uncompressed length. (a) What magnitude of force must you apply to hold the platform in this position? (b) How much \(additional\) work must you do to move the platform 0.200 m \(farther\), and what maximum force must you apply?
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