Problem 51 - University Physics with Modern Physics

Question

A mysterious rocket-propelled object of mass 45.0 kg is initially at rest in the middle of the horizontal, frictionless surface of an ice-covered lake. Then a force directed east and with magnitude \(F(t) =\) (16.8 N/s)\(t\) is applied. How far does the object travel in the first 5.00 s after the force is applied?

Step-by-Step Solution

Answer

The object travels 7.78 meters in the first 5 seconds.

Step 1: Understand the Problem

We need to determine how far an object of mass 45.0 kg travels on a frictionless surface when a time-dependent force is applied. The force acting on the object is given by \( F(t) = 16.8t \) N and acts over 5 seconds.

Step 2: Find the Expression for Acceleration

First, apply Newton's Second Law, \( F = ma \). The force function is \( F(t) = 16.8t \), and the mass \( m \) is 45.0 kg. Substitute these into the equation to find acceleration as a function of time: \[ a(t) = \frac{F(t)}{m} = \frac{16.8t}{45.0} \]}

Step 3: Integrate to Find Velocity Function

The acceleration \( a(t) \) is the derivative of velocity with respect to time. Integrate \( a(t) \) to find the velocity function. \[ v(t) = \int a(t) \, dt = \int \frac{16.8t}{45.0} \, dt = \frac{16.8}{90}t^2 + C \] Since the object starts from rest, \( v(0) = 0 \), so \( C = 0 \). Thus, \[ v(t) = \frac{16.8}{90}t^2 \]

Step 4: Integrate to Find Position Function

Velocity \( v(t) \) is the derivative of position \( x(t) \). Integrate the velocity function to find position. \[ x(t) = \int v(t) \, dt = \int \frac{16.8}{90}t^2 \, dt = \frac{16.8}{270}t^3 + C \]Since the object starts from the origin, \( x(0) = 0 \), so \( C = 0 \). Thus, \[ x(t) = \frac{16.8}{270}t^3 \]

Step 5: Calculate the Distance Travelled at \( t = 5 \) seconds

Substitute \( t = 5 \) seconds into the position equation to find the distance traveled.\[ x(5) = \frac{16.8}{270} \times 5^3 = \frac{16.8}{270} \times 125 \]Calculate the expression to find \( x(5) \) .

Step 6: Solve for Final Answer

Finally, computing the expression gives \( x(5) = \frac{16.8 \times 125}{270} = 7.78 \) meters. Thus, the object travels 7.78 meters in the first 5 seconds after the force is applied.

Key Concepts

Kinematics

Kinematics is the branch of physics that deals with the motion of objects. It focuses on concepts such as velocity, acceleration, and displacement without considering the causes of the motion. In the given problem, we explore how an object moves across a frictionless surface under the influence of a force that varies with time.

• Displacement: This is the overall change in position of an object. It is a critical variable in kinematics as we look to find out how far the object travels.
• Velocity: This is the rate of change of displacement. It's important to know how fast an object is moving, which starts from rest in this exercise.
• Acceleration: The rate of change of velocity. In this scenario, it's not constant due to the time-dependent force applied to the object.

To solve the exercise, the equation of motion for the object was formed using these kinematic concepts, requiring integration to find the expressions for velocity and displacement.

Integration

Integration is a mathematical technique used to find quantities like area, displacement, and volume. In physics, especially kinematics, it allows us to derive velocity and position from known rates of change.

• From Acceleration to Velocity: Once we have the acceleration as a function of time, we integrate it with respect to time (t) to determine the velocity function. This step involves solving the integral of the time-dependent acceleration formula.
• From Velocity to Position: Similarly, the velocity function, when integrated over time, gives us the position function. This helps us determine the total displacement of the object after a certain interval, in this case, after 5 seconds.
• Constants of Integration: Each time you integrate, a constant of integration (C) is introduced. Initial conditions, like starting from rest or the origin, help solve for these constants.

Through integration, we transition from having a force function to knowing how far the object travels when subjected to that force.

Time-Dependent Force

A time-dependent force is a force that changes with time. In our problem, the force exerted on the object is described by the equation \( F(t) = 16.8t \) N, indicating that the force increases linearly with time. This contrasts with constant forces, making the problem more dynamic.

• Understanding Time Variation: As the object experiences a force that grows as time progresses, its acceleration is not constant but also dependent on time.
• Impulse and Momentum: While not explicitly covered here, time-dependent forces also tie into concepts of impulse, which is the change in momentum. However, we focused on its role in altering velocity and producing displacement.
• Applying Newton's Second Law: This principle, \( F = ma \), is used to relate the time-dependent force to the acceleration of the object. Because the force changes with time, so does the acceleration.

Incorporating the concept of a time-dependent force shows how external influences changing with time affect an object's motion, which can be analyzed using kinematics and integration.