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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) The acceleration is approximately 4.31 m/s虏. (b) The crate travels 215.5 meters in 10 seconds. (c) Its speed at the end is 43.1 m/s.

Step by step solution

01

Calculate the Acceleration (Newton's Second Law)

To find the acceleration, we use Newton's second law of motion, which is given by \( F = ma \). Here, \( F = 140 \, \text{N} \) is the net force and \( m = 32.5 \, \text{kg} \) is the mass of the crate. Solving for acceleration \( a \):\[ a = \frac{F}{m} = \frac{140}{32.5} \approx 4.31 \, \text{m/s}^2 \].
02

Calculate the Distance Traveled (Kinematic Equation)

With the crate starting at rest (initial velocity \( u = 0 \)) and given time \( t = 10.0 \, \text{s}\), we use the kinematic equation: \[ s = ut + \frac{1}{2}at^2 \]. Plugging in the known values: \[ s = 0 + \frac{1}{2} \times 4.31 \times (10.0)^2 = \frac{1}{2} \times 4.31 \times 100 = 215.5 \, \text{m} \].
03

Calculate Final Speed (Kinematic Equation)

The final speed \( v \) after time \( t \) can be found using the equation \[ v = u + at \]. Since the initial speed \( u = 0 \), the equation simplifies to:\[ v = 0 + 4.31 \times 10.0 = 43.1 \, \text{m/s} \].

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Key Concepts

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

Understanding Kinematic Equations
When an object moves, kinematic equations help us predict its future motion based on its current state. These equations are crucial when an object experiences constant acceleration, like the crate in our problem. The most common kinematic equations relate acceleration, initial velocity, final velocity, distance covered, and time interval. They allow us to solve for unknown quantities if other variables are known.

For instance, to find the distance traveled by a crate starting from rest, we use:
  • Initial velocity, usually denoted as \( u \), is 0 if the object starts from rest.
  • The kinematic equation: \( s = ut + \frac{1}{2}at^2 \), where \( s \) is the distance, \( a \) is acceleration, and \( t \) is time.
By substituting the known values, we calculate how far the crate moves over time. Using these equations is like filling in pieces of a puzzle with math.
The Concept of Net Force
Newton's second law states that a net force acting on an object results in acceleration. When we say 'net force', we mean the total force after all individual forces on the object are combined. If multiple forces act in different directions, their vector sum or resultant force is the net force.

Think of net force as the driving factor that decides how quickly an object changes its velocity. This is why we use the formula \( F = ma \), which means force equals mass times acceleration.
  • Given a mass, if the net force increases, the acceleration increases proportionally.
  • Conversely, with the same net force, a larger mass will result in slower acceleration.
Understanding net force allows us to predict and manipulate the motion of objects.
Calculating Acceleration
Acceleration tells us how quickly the velocity of an object is changing. It's a crucial component of motion, especially when forces are involved. To calculate acceleration, you divide the net force applied to an object by its mass:

  • Formula: \( a = \frac{F}{m} \)
  • Where \( F \) is the net force, and \( m \) is the mass.
In our problem, the crate, with a mass of 32.5 kg, experiences a net force of 140 N. Using the formula, we find that the acceleration is approximately \( 4.31 \, \text{m/s}^2 \).

This acceleration describes how much faster the crate becomes every second, giving us insight into its changing speed as it moves across the floor. Understanding and calculating acceleration is fundamental to studying the motion in physics.

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Most popular questions from this chapter

CP A 4.80 -kg bucket of water is accelerated upward by a cord of negligible mass whose breaking strength is 75.0 \(\mathrm{N}\) . If the bucket starts from rest, what is the minimum time required to raise the bucket a vertical distance of 12.0 \(\mathrm{m}\) without breaking the cord?

Two crates, one with mass 4.00 \(\mathrm{kg}\) and the other with mass \(6.00 \mathrm{kg},\) sit on the frictionless surface of a frozen pond, connected by a light rope (Fig. P4.43). A woman wearing golf shoes (so she can get traction on the ice) pulls horizontally on the 6.00 -kg crate with a force \(F\) that gives the crate an acceleration of 2.50 \(\mathrm{m} / \mathrm{s}^{2} .\) (a) What is the acceleration of the 4.00 -kg crate? (b) Draw a free-body diagram for the 4.00 -kg crate. Use that diagram and Newton's second law to find the tension \(T\) in the rope that connects the two crates. (c) Draw a free-body diagram for the 6.00 -kg crate. What is the direction of the net force on the 6.00 -kg crate? Which is larger in magnitude, force \(T\) or force \(F ?(\mathrm{d})\) Use part \((\mathrm{c})\) and Newton's second law to calculate the magnitude of the force \(F .\)

A small car (mass 380 \(\mathrm{kg}\) ) is pushing a large truck (mass900 \(\mathrm{kg}\) ) due east on a level road. The car exerts a horizontal force of 1200 \(\mathrm{N}\) on the truck. What is the magnitude of the force that the truck exerts on the car?

World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \(\mathrm{m} / \mathrm{s}^{2}\) . How much horizontal force must a 55 -kg sprinter exert on the starting blocks during a start to produce this acceleration? Which body exerts the force that propels the sprinter: the blocks or the sprinter herself?

Two dogs pull horizontally on ropes attached to a post; the angle between the ropes is \(60.0^{\circ} .\) If dog \(A\) exerts a force of 270 \(\mathrm{N}\) and dog \(B\) exerts a force of \(300 \mathrm{N},\) find the magnitude of the resultant force and the angle it makes with dog \(A\) 's rope.
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