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Chapter 4: Problem 12

A crate with mass 32.5 \(\mathrm{kg}\) initially at rest on a warehouse floor is acted on by a net horizontal force of 140 \(\mathrm{N}\) . (a) What acceleration is produced? (b) How far does the crate travel in 10.0 \(\mathrm{s} ?\) (c) What is its speed at the end of 10.0 \(\mathrm{s} ?\)

Short Answer

Expert verified
a) 4.31 m/s虏, b) 215.5 m, c) 43.1 m/s.

Step by step solution

01

Understand the Problem

We are given a crate with a mass of 32.5 kg on which a net horizontal force of 140 N is acting. We need to find (a) the acceleration produced, (b) the distance traveled in 10.0 seconds, and (c) its speed at the end of 10.0 seconds.
02

Calculate the Acceleration (Part a)

To find the acceleration, use Newton's second law, which states \( F = m \, a \), where \( F \) is the force applied, \( m \) is the mass, and \( a \) is the acceleration. \( a = \frac{F}{m} = \frac{140}{32.5} \). Calculating gives \( a \approx 4.31 \, \text{m/s}^2 \).
03

Calculate the Distance Traveled in 10 Seconds (Part b)

Use the formula for distance when starting from rest: \( d = \frac{1}{2} a t^2 \). With \( a = 4.31 \, \text{m/s}^2 \) and \( t = 10.0 \, \text{s} \), we find \( d = \frac{1}{2} \times 4.31 \times 10^2 = 215.5 \, \text{m} \).
04

Calculate the Speed at the End of 10 Seconds (Part c)

Use the formula \( v = u + at \) where \( u \) is the initial speed (0 in this case), \( a \) is the acceleration, and \( t \) is time. Calculating, \( v = 0 + 4.31 \times 10 = 43.1 \, \text{m/s} \).

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Key Concepts

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

Acceleration Calculation
Calculating acceleration is an essential concept in understanding how forces affect motion. Newton's second law of motion provides the key formula: \( F = m \cdot a \), where \( F \) is the force applied, \( m \) is the mass, and \( a \) is the acceleration. When a force acts on an object, it changes the object's motion, producing acceleration. The formula rearranges to \( a = \frac{F}{m} \), enabling us to find acceleration when force and mass are known. In our example, the crate experiences a force of 140 N. With a mass of 32.5 kg, using \( a = \frac{140}{32.5} \) results in an acceleration \( a \approx 4.31 \text{ m/s}^2 \).Understanding this calculation helps predict how quickly an object speeds up or slows down when subjected to external forces.
Distance Traveled
The distance an object travels over time while starting from rest can be determined using the formula \( d = \frac{1}{2} a t^2 \). This equation is derived from the basic principles of kinematic motion equations and helps us find out how far an object has moved under constant acceleration.When tackling problems of this nature, it's crucial to know:
  • Initial speed \( (u) \)
  • Acceleration \( (a) \)
  • Time \( (t) \) elapsed
For the crate problem, starting from rest means initial velocity \( u = 0 \). With an acceleration \( a \approx 4.31 \text{ m/s}^2 \) and time \( t = 10 \text{ s} \), the distance covered is \( d \approx 215.5 \text{ m} \).Using such equations, understanding how different variables affect distance can help you predict real-world motion scenarios effectively.
Final Velocity
To determine an object's speed at a certain point in time when it begins from rest with constant acceleration, we employ the formula \( v = u + at \). This particular equation is incredibly useful to predict how fast an object will be going after a set period of time.In our scenario, we make use of the following information:
  • Initial velocity \( u = 0 \)
  • Calculated acceleration \( a \approx 4.31 \text{ m/s}^2 \)
  • Time \( t = 10 \text{ s} \)
Thus, substituting the known values gives us \( v = 0 + 4.31 \times 10 = 43.1 \text{ m/s} \). The final velocity represents how fast the crate is moving after 10 seconds.Grasping this concept is key in understanding how consistent acceleration influences an object's speed over time.

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Most popular questions from this chapter

An oil tanker's engines have broken down, and the wind is blowing the tanker straight toward a recf at a constant spocd of 1.5 \(\mathrm{m} / \mathrm{s}(\text { Fig. } 4.37) .\) When the tanker is 500 \(\mathrm{m}\) m from the reef, the wind dies down just as the engineer gets the engines going again. The rudder is stuck, so the only choice is to try to accelerate straight backward away from the reef. The mass of the tanker and cargo is \(3.6 \times 10^{7} \mathrm{kg}\) , and the engines produce a net horizontal force of \(8.0 \times 10^{4} \mathrm{N}\) on the tanker. Will the ship hit the reef? If it does, will the oil be safe? The hull can withstand an impact at a speed of 0.2 \(\mathrm{m} / \mathrm{s}\) or less. You can ignore the retarding force of the water on the tanker's hull.

If a net horizontal force of 132 \(\mathrm{N}\) is applied to a person with mass 60 \(\mathrm{kg}\) who is resting on the edge of a swimming pool, what horizontal acceleration is produced?

A \(4.80-\mathrm{kg}\) bucket of water is accelerated upward by a cord of negligible mass whose breaking strength is 75.0 \(\mathrm{N}\) . (a) Draw the free-body force diagram for the bucket. In terms of the forces on your diagram, what is the net force on the bucket? (b) Apply Newton's second law to the bucket and find the maximum upward acceleration that can be given to the bucket without breaking the cord.

If we know \(F(t),\) the force as a function of time, for straight-line motion, Newton's second law gives us \(a(t),\) the acceleration as a function of time. We can then integrate \(a(t)\) to find \(v(t)\) and \(x(t)\) . However, suppose we know \(F(v)\) instead. (a) The net force on a body moving along the \(x\) -axis equals \(-C v^{2} .\) Use Newton's second law written as \(\Sigma F=m d v / d t\) and two integrations to show that \(x-x_{0}=(m / C) \ln \left(v_{0} / v\right) .\) (b) Show that Newton's second law can be written as \(\Sigma F=m v d v / d x .\) Derive the same expression as in part (a) using this form of the second law and one integration.

An astronaut is tethered by a strong cable to a spacecraft. The astronaut and her spacesuit have a total mass of 105 \(\mathrm{kg}\) , while the mass of the cable is negligible. The mass of the spacecraft is \(9.05 \times 10^{4} \mathrm{kg}\) . The spacecraft is from any large astronomical bodies, so we can ignore the gravitational forces on it and the astronaut. We also assume that both the spacecraft and the astronaut are initially at rest in an inertial reference frame. The astronaut then pulls on the cable with a force of 80.0 \(\mathrm{N}\) . (a) What force bodies, so we can ignore the gravitational forces on it and the astronaut. We also assume that both the spacecraft and the astronaut are initially at rest in an inertial reference frame. The astronaut then pulls on the cable with a force of 80.0 \(\mathrm{N}\) . (a) What force does the cable exert on the astronaut? (b) Since \(\Sigma \overrightarrow{\boldsymbol{F}}=m \overrightarrow{\mathbf{a}},\) how can a "massless" \((m=0)\) cable exert a force? (c) What is the astronaut's acceleration? (d) What force does the cable exert on the spacecraft? (e) What is the acceleration of the spacecraft?
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