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In outer space, a constant net force of magnitude 140 \(\mathrm{N}\) is exerted on a 32.5 \(\mathrm{kg}\) probe initially at rest. (a) What acceleration does this force produce? (b) How far does the probe travel in 10.0 \(\mathrm{s} ?\)

Short Answer

Expert verified
(a) The acceleration is approximately 4.31 m/s². (b) The probe travels approximately 215.5 m in 10 seconds.

Step by step solution

01

Identify the Given Values

We are given a net force \( F = 140 \ \text{N} \) and the mass of the probe \( m = 32.5 \ \text{kg} \). The probe is initially at rest, so its initial velocity \( v_0 = 0 \ \text{m/s} \). We also need to find the acceleration \( a \) and the distance \( s \) traveled in \( t = 10.0 \ \text{s} \).
02

Calculate Acceleration Using Newton's Second Law

According to Newton's second law, force is expressed as \( F = ma \). We can solve for acceleration, \( a \), using the formula \( a = \frac{F}{m} \). Substitute the given values to get: \[ a = \frac{140 \ \text{N}}{32.5 \ \text{kg}} \].
03

Perform Acceleration Calculation

Solve the equation: \[ a = \frac{140}{32.5} \approx 4.31 \ \text{m/s}^2 \]. This is the acceleration produced by the force on the probe.
04

Use Kinematics to Find the Distance Traveled

We use the kinematic equation for motion under constant acceleration: \[ s = v_0 t + \frac{1}{2} a t^2 \]. Given that the initial velocity \( v_0 = 0 \), this simplifies to: \[ s = \frac{1}{2} a t^2 \].
05

Substitute Values to Calculate Distance

Substitute \( a = 4.31 \ \text{m/s}^2 \) and \( t = 10.0 \ \text{s} \) into the equation: \[ s = \frac{1}{2} \times 4.31 \times 10^2 \].
06

Perform Distance Calculation

Calculate the expression: \[ s = \frac{1}{2} \times 4.31 \times 100 \approx 215.5 \ \text{m} \]. This is the distance the probe travels in 10 seconds.

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Key Concepts

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

Kinematic Equations
Kinematic equations are a fundamental part of motion in physics, especially when dealing with constant acceleration. These equations allow us to make predictions about an object's movement based on its initial conditions, such as velocity and position, along with its acceleration.

In the exercise, we make use of the equation \( s = v_0 t + \frac{1}{2} a t^2 \) to calculate the distance an object travels under constant acceleration. It's crucial to understand each part of this equation. Here are the components:

  • \(s\) - the distance traveled.
  • \(v_0\) - the initial velocity, which is 0 in the exercise.
  • \(t\) - the time for which the motion occurs.
  • \(a\) - the constant acceleration.


When the object starts from rest, the initial velocity \(v_0\) is zero, simplifying the equation to \( s = \frac{1}{2} a t^2 \). This scenario portrays a straightforward calculation for objects under constant acceleration. By substituting known values of acceleration and time, you can compute the distance impelled by a constant force.
Constant Acceleration
Constant acceleration refers to a scenario where the acceleration of an object remains unchanged throughout the time period under consideration. It plays a central role when analyzing motion with the help of kinematic equations.

In the example provided, the acceleration is determined using Newton's second law as follows: \( a = \frac{F}{m} \). Here, the net force \( F \) and mass \( m \) remain constant, resulting in a steady acceleration value of approximately \( 4.31 \, \text{m/s}^2 \).

This constancy means the object gains speed uniformly over time, making it easy to predict the probe's motion using kinematic equations. Constant acceleration simplifies the calculations as it ensures that the velocity changes by the same amount in equal intervals of time.

Understanding constant acceleration helps in resolving numerous real-world physics problems, similar to the probe in space. It provides a reliable framework for predicting motion accurately over specified time durations.
Force and Motion
Force and motion are interconnected through Newton's second law of motion, which bridges these physical concepts. This law states that a force applied to an object causes it to accelerate in the direction of the force, and it can be mathematically represented as \( F = ma \). This formula is pivotal when solving problems related to motion and was utilized in the exercise to find acceleration first.

Here are some key points to remember:
  • A larger force results in greater acceleration for a given mass.
  • For the same force, a larger mass will result in a smaller acceleration.
  • The acceleration occurs in the direction of the net force.


In the exercise, when a constant force of \( 140 \, \text{N} \) acts on a \( 32.5 \, \text{kg} \) probe, it leads to a calculated acceleration of \( 4.31 \, \text{m/s}^2 \). This relationship illustrates how force precipitates motion in a predictable and calculable manner.

Through mastering the connection between force and motion, you can understand how various forces around us affect objects, an essential concept in both fundamental physics education and practical applications.

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Most popular questions from this chapter

A warehouse worker pushes a crate along the floor, as shown in Figure \(4.33,\) by a force of 10 \(\mathrm{N}\) that points downward at an angle of \(45^{\circ}\) below the horizontal. Find the horizontal and vertical components of the push.

A dock worker 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 the first 5.00 \(\mathrm{s}\) s. What is the mass of the block of ice?

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Superman throws a 2400 -N boulder at an adversary. What horizontal force must Superman apply to the boulder to give it a horizontal acceleration of 12.0 \(\mathrm{m} / \mathrm{s}^{2} ?\)

You drag a heavy box along a rough horizontal floor by a horizontal rope. Identify the reaction force to each of the following forces: (a) the pull of the rope on the box, (b) the friction force on the box, (c) the normal force on the box, and (d) the weight of the box.

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