/*! 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 10 A dockworker applies a constant ... [FREE SOLUTION] | 91Ó°ÊÓ

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A dockworker applies a constant horizontal force of 80.0 \(\mathrm{N}\) to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 \(\mathrm{m}\) in 5.00 \(\mathrm{s}\) . (a) What is the mass of the block of ice? (b) If the worker stops pushing at the end of 5.00 \(\mathrm{s}\) s, how far does the block move in the next 5.00 \(\mathrm{s?}\)

Short Answer

Expert verified
(a) The mass of the block is approximately 90.91 kg. (b) The block moves 22.0 m after the force ceases.

Step by step solution

01

Determine Acceleration

Use the formula for uniform acceleration: \( s = ut + \frac{1}{2}at^2 \) where \( s = 11.0 \, \text{m} \), \( u = 0 \, \text{m/s} \) (since it starts from rest), and \( t = 5.00 \, \text{s} \). Simplifying, we get \( 11.0 = \frac{1}{2}a(5.00^2) \). Solving for \( a \), \( a = \frac{2 \times 11.0}{5.00^2} = 0.88 \, \text{m/s}^2 \).
02

Find Mass Using Force Equation

Use Newton's second law of motion, \( F = ma \). The force \( F = 80.0 \, \text{N} \) and the acceleration \( a = 0.88 \, \text{m/s}^2 \). Now solve for mass \( m \): \( m = \frac{F}{a} = \frac{80.0}{0.88} \approx 90.91 \, \text{kg} \).
03

Find Final Velocity after 5 Seconds

Use the formula \( v = u + at \), where \( u = 0 \, \text{m/s} \), \( a = 0.88 \, \text{m/s}^2 \), and \( t = 5.00 \, \text{s} \). Solving gives \( v = 0 + (0.88)(5.00) = 4.4 \, \text{m/s} \).
04

Determine Distance After Force Ceases

Once the worker stops pushing, the block will move uniformly with the final velocity from step 3. Use \( s = vt \). So, \( s = (4.4)(5.00) = 22.0 \, \text{m} \).

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Key Concepts

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

Uniform Acceleration
Uniform acceleration occurs when an object's velocity changes at a constant rate. It's a key concept in understanding motion, especially in problems involving kinematic equations. In our exercise, the block of ice moved with uniform acceleration because the dockworker applied a constant force, and there was negligible frictional resistance.

When you know an object started from rest, the initial velocity (\(u\)) equals zero. The kinematic equation that relates distance (\(s\)), initial velocity (\(u\)), time (\(t\)), and acceleration (\(a\)) is:
  • \[ s = ut + \frac{1}{2}at^2 \]
In this equation:
  • \( s \) is the distance covered,
  • \( u \) is the initial velocity (0 m/s in this case),
  • \( t \) is the time,
  • \( a \) is the acceleration.
By substituting the given values and solving, you can find the acceleration of the block. This underlying principle of constant acceleration allows us to predict how an object will behave over time when acted upon by a steady force.
Force and Motion
Newton's Second Law of Motion is essential for understanding the relationship between force, mass, and acceleration. It states that the force applied to an object is equal to the mass of the object multiplied by its acceleration, expressed by the formula:
  • \[ F = ma \]
In our scenario, the dockworker applied a force of 80 \( \text{N} \) to the block of ice. Using the acceleration previously calculated, you can rearrange the formula to solve for the mass (\(m\)). This becomes:
  • \[ m = \frac{F}{a} \]
This formula allows us to understand how an object's speed changes due to the application of a force. As the force is constant and does not change over time in our case, the mass of the block is simply found by dividing the force by the calculated acceleration of 0.88 \( \text{m/s}^2 \). Understanding force and motion helps explain how different magnitudes of force and mass affect acceleration.
Kinematics
Kinematics is the branch of physics that describes the motion of objects without considering the forces that cause this motion. It involves analyzing movement through equations that relate velocity, acceleration, time, and displacement.

After determining the block's acceleration and applying Newton's second law, we use the principles of kinematics to learn more about the motion of our block of ice. First, we calculate the final velocity after 5 seconds using:
  • \[ v = u + at \]
Here,\( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is time.
Knowing the block moves uniformly once the worker stops pushing, we calculate how far it travels with its final velocity. The distance is then computed as:
  • \[ s = vt \]
In our case, this simplifies calculating the distance traveled when no further force is applied. Applying kinematic equations thus provides clarity on the continuous nature of movement under different conditions.

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Most popular questions from this chapter

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