/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 You raise a bucket of water from... [FREE SOLUTION] | 91Ó°ÊÓ

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You raise a bucket of water from the bottom of a deep well. If your power output is \(108 \mathrm{W}\), and the mass of the bucket and the water in it is \(5.00 \mathrm{kg}\), with what speed can you raise the bucket? Ignore the weight of the rope.

Short Answer

Expert verified
2.20 \(\mathrm{ms^{-1}}\)

Step by step solution

01

Understanding the Relationship Between Power, Force, and Velocity

Power is defined as the rate at which work is done. For this scenario, power can be calculated using the equation: \( P = F \cdot v \), where \( P \) is the power, \( F \) is the force exerted, and \( v \) is the velocity (or speed) at which the bucket is raised.
02

Calculate the Force Required to Lift the Bucket

The force required to lift the bucket is equal to the gravitational force acting on it. This force can be calculated using \( F = m \cdot g \), where \( m = 5.00 \mathrm{kg} \) is the mass of the bucket and water, and \( g = 9.81 \mathrm{ms^{-2}} \) is the acceleration due to gravity. Thus, \( F = 5.00 \times 9.81 = 49.05 \mathrm{N} \).
03

Solve for Velocity Using the Power Formula

We need to rearrange the power formula to solve for velocity: \( v = \frac{P}{F} \). Substituting the values we have, \( v = \frac{108}{49.05} \approx 2.20 \mathrm{ms^{-1}} \).
04

Conclusion

The velocity, or speed, at which you can raise the bucket with a power output of \(108 \mathrm{W}\) is approximately \(2.20 \mathrm{ms^{-1}}\).

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Key Concepts

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

gravitational force
Gravitational force is a fundamental force of nature that acts between two masses. In simpler terms, it’s the force that gives us weight and keeps objects grounded on Earth. Every object with mass has a gravitational pull, but the Earth’s large mass makes it the dominant force we experience daily. When you hold up a bucket of water, the gravitational force pulls it downwards toward the Earth’s center. This is what we commonly understand as the weight of the object.

To calculate the gravitational force acting on an object, you use the formula:
\( F = m \cdot g \)
where:
  • \( F \) is the gravitational force (in Newtons)
  • \( m \) is the mass of the object (in kilograms)
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \mathrm{ms^{-2}} \) on Earth

In our bucket scenario, the weight or gravitational force acting on the bucket is found by multiplying its mass by Earth's gravitational pull, resulting in \( 49.05 \mathrm{N} \). This force must be overcome to lift the bucket.
power formula
The power formula is a fundamental concept in physics, used to describe how quickly work is done or energy is transferred. Power is essentially the rate of doing work. You can think of it as how much energy is used or how much work is executed in a particular period of time.

The formula for calculating power is given by:
\( P = F \cdot v \)
where:
  • \( P \) represents power (in watts)
  • \( F \) represents the force applied (in Newtons)
  • \( v \) is the velocity at which the work is done (in meters per second)

In this exercise, the power output was specified as \( 108 \mathrm{W} \) when lifting a bucket. By rearranging this equation, as shown in the solution, you can solve for velocity by dividing power by force: \( v = \frac{P}{F} \).

This means to lift the bucket, you are converting the power into lifting speed. If force remains constant, a higher power means the work is done faster or with a greater speed.
acceleration due to gravity
Acceleration due to gravity is a constant value that represents the rate at which an object will increase its speed as it falls freely towards the Earth. On Earth, the average value of this acceleration is \( 9.81 \mathrm{ms^{-2}} \). This constant is vital for understanding how forces like gravity influence the motion of objects.

When an object is in free fall, gravitational acceleration is the only force acting on it, causing it to speed up. This concept works similarly when you lift objects; you have to exert a force against the acceleration due to gravity to move them upward.

With this understanding, the gravitational force in any physical scenario, like raising a bucket out of a well, is directly related to this principle of acceleration due to gravity. Therefore, the rate of acceleration influences how much force you need to apply to lift an object effectively.

Understanding acceleration due to gravity helps in making calculations involving lifting and dropping objects, as seen in the exercise where it was used to calculate the force required to lift the bucket.

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Most popular questions from this chapter

You throw a glove straight upward to celebrate a victory. Its initial kinetic energy is \(K\) and it reaches a maximum height \(h\) What is the kinetic energy of the glove when it is at the height \(h / 2 ?\)

IP A sled with a mass of \(5.80 \mathrm{kg}\) is pulled along the ground through a displacement given by \(\overrightarrow{\mathrm{d}}=(4.55 \mathrm{m}) \hat{\mathrm{x}}\). (Let the \(x\) axis be horizontal and the \(y\) axis be vertical.) (a) How much work is done on the sled when the force acting on it is \(\overrightarrow{\mathbf{F}}=(2.89 \mathrm{N}) \hat{\mathrm{x}}+\) \((0.131 \mathrm{N}) \hat{\mathrm{y}} ?\) (b) How much work is done on the sled when the force acting on it is \(\overrightarrow{\mathbf{F}}=(2.89 \mathrm{N}) \hat{\mathrm{x}}+(0.231 \mathrm{N}) \hat{\mathrm{y}} ?\) (c) If the mass of the sled is increased, does the work done by the forces in parts (a) and (b) increase, decrease, or stay the same? Explain.

IP A kayaker paddles with a power output of \(50.0 \mathrm{W}\) to maintain a steady speed of \(1.50 \mathrm{m} / \mathrm{s}\). (a) Calculate the resistive force exerted by the water on the kayak. (b) If the kayaker doubles her power output, and the resistive force due to the water remains the same, by what factor does the kayaker's speed change?

C E Predict/Explain The work \(W_{0}\) accelerates a car from 0 to \(50 \mathrm{km} / \mathrm{h} .\) (a) Is the work required to accelerate the car from \(50 \mathrm{km} / \mathrm{h}\) to \(150 \mathrm{km} / \mathrm{h}\) equal to \(2 \mathrm{W}_{0}, 3 \mathrm{W}_{0}, 8 \mathrm{W}_{0},\) or \(9 \mathrm{W}_{0} ?\) (b) Choose the best explanation from among the following: I. The work to accelerate the car depends on the speed squared. II. The final speed is three times the speed that was produced by the work \(W_{0}\). III. The increase in speed from \(50 \mathrm{km} / \mathrm{h}\) to \(150 \mathrm{km} / \mathrm{h}\) is twice the increase in speed from 0 to \(50 \mathrm{km} / \mathrm{h}\).

IP A 9.50-g bullet has a speed of \(1.30 \mathrm{km} / \mathrm{s}\). (a) What is its \(k i\) netic energy in joules? (b) What is the bullet's kinetic energy if its speed is halved? \((\mathrm{c})\) If its speed is doubled?

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