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Chapter 8: Problem 62

\(\begin{array}{l}{\text { (I) How long will it take a } 1750-\mathrm{W} \text { motor to lift a } 335 \text { -kg }} \\ {\text { piano to a sixth-story window } 16.0 \mathrm{m} \text { above? }}\end{array}\)

Short Answer

Expert verified
The motor will take 30.0 seconds to lift the piano.

Step by step solution

01

Identify the Given Values

The power of the motor is given as \( P = 1750 \, \text{W} \) and the mass of the piano is \( m = 335 \, \text{kg} \). The height the piano needs to be lifted is \( h = 16.0 \, \text{m} \).
02

Determine the Gravitational Force

The gravitational force acting on the piano can be calculated using the formula \( F_g = m \cdot g \), where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Thus, \( F_g = 335 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 3283 \, \text{N} \).
03

Calculate the Work Done

Work done by the motor when lifting the piano can be calculated using the formula \( W = F_g \times h \). Substituting the known values, \( W = 3283 \, \text{N} \times 16.0 \, \text{m} = 52528 \, \text{J} \).
04

Apply the Formula for Power and Time

Power can be related to work and time using the formula \( P = \frac{W}{t} \). We can rearrange this formula to find time: \( t = \frac{W}{P} \).
05

Calculate the Time Required

Substitute the known values into the rearranged formula: \( t = \frac{52528 \, \text{J}}{1750 \, \text{W}} = 30.0 \, \text{s} \).

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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 the force with which the Earth pulls objects towards its center. It's what gives objects weight. For any object near the Earth's surface, the gravitational force can be calculated using the formula:
  • \( F_g = m \cdot g \)
where:
  • \( F_g \) is the gravitational force in newtons (N),
  • \( m \) is the mass of the object in kilograms (kg), and
  • \( g \) is the acceleration due to gravity, which is approximately \( 9.8 \, \text{m/s}^2 \).
For the piano example, we plug in the mass \( m = 335 \, \text{kg} \) to get:
  • \( F_g = 335 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 3283 \, \text{N} \)
This tells us that the piano experiences a gravitational force of 3283 newtons downward.
Work Done
Work is done when a force causes an object to move in the direction of the force. The formula to find the work done is:
  • \( W = F \times d \)
where:
  • \( W \) is the work done in joules (J),
  • \( F \) is the force applied in newtons (N), and
  • \( d \) is the distance the object moves in meters (m).
For the piano, which is being lifted, the force is the gravitational force, and the distance is how high it is lifted. So, using:
  • Gravitational force \( F_g = 3283 \, \text{N} \)
  • Height \( h = 16.0 \, \text{m} \)
The work done is:
  • \( W = 3283 \, \text{N} \times 16.0 \, \text{m} = 52528 \, \text{J} \)
This means the motor does 52,528 joules of work to lift the piano.
Power Formula
Power is a measure of how quickly work is done. It is calculated using the formula:
  • \( P = \frac{W}{t} \)
where:
  • \( P \) is power in watts (W),
  • \( W \) is work done in joules (J), and
  • \( t \) is the time in seconds (s) over which the work is done.
In this exercise, we're given that the motor's power is \( 1750 \, \text{W} \). Power tells us how much work can be done in a certain amount of time and the capability of the motor to lift the piano as quickly as possible.
Time Calculation
Time calculation in this context tells us how long it takes for the motor to perform the work of lifting the piano. The formula relating power, work, and time is rearranged to solve for time:
  • \( t = \frac{W}{P} \)
By substituting in the known values:
  • \( W = 52528 \, \text{J} \)
  • \( P = 1750 \, \text{W} \)
The calculation for time becomes:
  • \( t = \frac{52528}{1750} = 30.0 \, \text{s} \)
This means it will take 30 seconds for the motor to lift the piano to the sixth floor. Understanding this formula helps us appreciate how changes in power or work affect the time required to complete a task.

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Most popular questions from this chapter

(II) A 96-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of \(350 \mathrm{~N}\). For the first \(15 \mathrm{~m}\) the floor is frictionless, and for the next \(15 \mathrm{~m}\) the coefficient of friction is \(0.25 .\) What is the final speed of the crate?

You drop a ball from a height of \(2.0 \mathrm{m},\) and it bounces back to a height of 1.5 \(\mathrm{m}\) (a) What fraction of its initial energy is lost during the bounce? (b) What is the ball's speed just before and just after the bounce? (c) Where did the energy go?

(II) A sled is initially given a shove up a frictionless \(23.0^{\circ}\) incline. It reaches a maximum vertical height \(1.12 \mathrm{~m}\) higher than where it started. What was its initial speed?

(1I) A 56 -kg skier starts from rest at the top of a 1200 -m- long trail which drops a total of 230 m from top to bottom. At the bottom, the skier is moving 11.0 \(\mathrm{m} / \mathrm{s} .\) How much. energy was dissignated by friction?

(II) A \(145-\mathrm{g}\) baseball is dropped from a tree \(14.0 \mathrm{~m}\) above the ground. ( \(a\) ) With what speed would it hit the ground if air resistance could be ignored? \((b)\) If it actually hits the ground with a speed of \(8.00 \mathrm{~m} / \mathrm{s}\), what is the average force of air resistance exerted on it?
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