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Chapter 9: Problem 58

Object Accelerates Horizontally An initially stationary \(2.0 \mathrm{~kg}\) object accelerates horizontally and uniformly to a speed of \(10 \mathrm{~m} / \mathrm{s}\) in \(3.0 \mathrm{~s}\). (a) In that \(3.0 \mathrm{~s}\) interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?

Short Answer

Expert verified
The work done is 99.9 J. The instantaneous power at the end of the interval is 66.6 W, and at the end of the first half is 33.3 W.

Step by step solution

01

Determine the acceleration

First, calculate the acceleration using the formula: \[ a = \frac{v_f - v_i}{t} \] where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( t \) is the time interval. Given \( v_f = 10 \mathrm{~m/s} \), \( v_i = 0 \mathrm{~m/s} \), and \( t = 3.0 \mathrm{~s} \): \[ a = \frac{10 \mathrm{~m/s} - 0}{3.0 \mathrm{~s}} = \frac{10}{3.0} \mathrm{~m/s^2} = 3.33 \mathrm{~m/s^2} \]
02

Calculate the force

Using Newton's second law, calculate the force: \[ F = m \times a \] where \( m \) is the mass and \( a \) is the acceleration. Given \( m = 2.0 \mathrm{~kg} \) and \( a = 3.33 \mathrm{~m/s^2} \): \[ F = 2.0 \mathrm{~kg} \times 3.33 \mathrm{~m/s^2} = 6.66 \mathrm{~N} \]
03

Calculate the work done

Work done by the force is given by: \[ W = F \times d \times \cos(\theta) \] Since the force and displacement are in the same direction, \( \theta = 0 \) and \( \cos(0) = 1 \). We need displacement \( d \), calculated as: \[ d = \frac{1}{2} a t^2 \] Given \( a = 3.33 \mathrm{~m/s^2} \) and \( t = 3.0 \mathrm{~s} \): \[ d = \frac{1}{2} \times 3.33 \mathrm{~m/s^2} \times (3.0 \mathrm{~s})^2 = 15 \mathrm{~m} \] Thus, the work done: \[ W = 6.66 \mathrm{~N} \times 15 \mathrm{~m} = 99.9 \mathrm{~J} \]
04

Calculate instantaneous power at the end of the interval

Instantaneous power at the end of the interval is: \[ P = F \times v \] where \( v \) is the velocity at the end of the interval. Given \( F = 6.66 \mathrm{~N} \) and \( v = 10 \mathrm{~m/s} \): \[ P = 6.66 \mathrm{~N} \times 10 \mathrm{~m/s} = 66.6 \mathrm{~W} \]
05

Calculate instantaneous power at the end of the first half of the interval

First, find the velocity at the end of 1.5 seconds (half the interval): \[ v_{1.5} = v_i + a \times t = 0 + 3.33 \mathrm{~m/s^2} \times 1.5 \mathrm{~s} = 5 \mathrm{~m/s} \] Then, calculate the power: \[ P = F \times v = 6.66 \mathrm{~N} \times 5 \mathrm{~m/s} = 33.3 \mathrm{~W} \]

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Key Concepts

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

Newton's Second Law
Newton's Second Law of Motion is fundamental in understanding how forces impact the motion of objects. This law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula is expressed as:

\( F = ma \)

Where:
  • \( F \) represents the net force applied
  • \( m \) is the mass of the object
  • \( a \) is the acceleration produced
In our exercise, a 2.0 kg object accelerates to 10 m/s in 3 seconds. Using Newton's Second Law, we calculate the force:

  • First, find the acceleration: \( a = \frac{10 \, \mathrm{m/s} - 0}{3 \, \mathrm{s}} = 3.33 \, \mathrm{m/s^2} \)
  • Then, use the law to find the force: \( F = 2.0 \, \mathrm{kg} \times 3.33 \, \mathrm{m/s^2} = 6.66 \, \mathrm{N} \)
Understanding this law helps explain how and why an object's velocity changes in response to the force applied.
Work and Energy
Work and energy are critical concepts in physics that describe how forces cause motion and transfer energy. Work is done when a force moves an object over a distance, and it's calculated using the formula:

\( W = F \, d \, \cos(\theta) \)

In our scenario, the force and displacement are in the same direction, simplifying our formula since \( \cos(0) = 1 \). We find the displacement first as:

  • \( d = \frac{1}{2} \, a \, t^2 \)
  • \( d = \frac{1}{2} \times 3.33 \, \mathrm{m/s^2} \times (3 \, \mathrm{s})^2 = 15 \, \mathrm{m} \)
Then, the work done is:

  • \( W = 6.66 \, \mathrm{N} \times 15 \, \mathrm{m} = 99.9 \, \mathrm{J} \)
Energy is often discussed in the context of work because work results in energy transfer. Here, the object's kinetic energy increases as work is done on it, leading to its final speed of 10 m/s.
Instantaneous Power
Power is the rate at which work is done or energy is transferred, and it can be instantaneous or average. Instantaneous power measures how much work is done at a specific moment in time. It is calculated with the formula:

\( P = F \, v \)

Where:
  • \( P \) is the instantaneous power
  • \( F \) is the force applied
  • \( v \) is the velocity at that instant
At the end of the 3.0-second interval, the velocity of the object is 10 m/s:

  • \( P = 6.66 \, \mathrm{N} \times 10 \, \mathrm{m/s} = 66.6 \, \mathrm{W} \)
At the end of the first half of the interval (1.5 seconds), we find the velocity is 5 m/s:

  • \( v_{1.5} = 0 + 3.33 \, \mathrm{m/s^2} \times 1.5 \, \mathrm{s} = 5 \, \mathrm{m/s} \)
  • \( P = 6.66 \, \mathrm{N} \times 5 \, \mathrm{m/s} = 33.3 \, \mathrm{W} \)
This shows how the instantaneous power changes with the velocity of the object, highlighting the dynamics of force and speed in the energy transfer process.

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Most popular questions from this chapter

A Proton is Accelerated A proton (mass \(m=1.67 \times 10^{-27} \mathrm{~kg}\) ) is being accelerated along a straight line at \(3.6 \times 10^{13} \mathrm{~m} / \mathrm{s}^{2}\) in a machine. If the proton has an initial speed of \(2.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) and travels \(3.5 \mathrm{~cm}\), what then is (a) its speed and (b) the increase in its kinetic energy?

Rope Tow A skier is pulled by a tow rope up a frictionless ski slope that makes an angle of \(12^{\circ}\) with the horizontal. The rope moves parallel to the slope with a constant speed of \(1.0 \mathrm{~m} / \mathrm{s}\). The force of the rope does \(900 \mathrm{~J}\) of work on the skier as the skier moves a distance of \(8.0 \mathrm{~m}\) up the incline. (a) If the rope moved with a constant speed of \(2.0 \mathrm{~m} / \mathrm{s}\), how much work would the force of the rope do on the skier as the skier moved a distance of \(8.0 \mathrm{~m}\) up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) \(1.0 \mathrm{~m} / \mathrm{s}\) and \((\mathrm{c})\) \(2.0 \mathrm{~m} / \mathrm{s} ?\)

A Sprinter A sprinter who weighs \(670 \mathrm{~N}\) runs the first \(7.0 \mathrm{~m}\) of a race in \(1.6 \mathrm{~s}\), starting from rest and accelerating uniformly. What are the sprinter's (a) speed and (b) kinetic energy at the end of the \(1.6 \mathrm{~s} ?\) (c) What average power does the sprinter generate during the 1.6 s interval?

Block Pulled at Constant Speed A \(100 \mathrm{~kg}\) block is pulled at a constant speed of \(5.0 \mathrm{~m} / \mathrm{s}\) across a horizontal floor by an applied force of \(122 \mathrm{~N}\) directed \(37^{\circ}\) above the horizontal. What is the rate at which the force does work on the block?

Block Dropped on a Spring A \(250 \mathrm{~g}\) block is dropped onto a relaxed vertical spring that has a spring constant of \(k=\) \(2.5 \mathrm{~N} / \mathrm{cm}\) (Fig. \(9-29)\). The block becomes attached to the spring and compresses the spring \(12 \mathrm{~cm}\) before turning around. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?
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