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Chapter 4: Problem 29

\(\Lambda\) force \(F=\left(10 t-t^{2}\right) \mathrm{N}\) acts on a body of mass \(100 \mathrm{~kg}\). Ilow far will it move, starting from rest, before it starts to reverse its direction.

Short Answer

Expert verified
The body will move 8.42 meters before it starts to reverse direction.

Step by step solution

01

Determine when the force becomes zero

To find when the body starts to reverse its direction, we need to find when the force, and thus the acceleration, becomes zero. Any further time after this will result in negative acceleration. Set the force equation equal to zero and solve for time:\[ F = 10t - t^2 = 0 \]Solve it to find the critical times it:\[ t(10 - t) = 0 \]\[ t = 0 \text{ or } t = 10 \]So, at \( t = 10 \text{ seconds} \), the force becomes zero.
02

Calculate acceleration as a function of time

Since force \( F = ma \), where \( m \) is the mass and \( a \) is the acceleration, we can find acceleration as a function of time:\[ a(t) = \frac{F}{m} = \frac{10t - t^2}{100} \]This simplifies to:\[ a(t) = 0.1t - 0.01t^2 \]
03

Integrate acceleration to find velocity

For the body starting from rest, integrate the acceleration function to find velocity as a function of time:\[ v(t) = \int a(t) \, dt = \int (0.1t - 0.01t^2) \, dt \]\[ v(t) = \left( 0.05t^2 - 0.0033t^3 \right) + C \]Since the body starts from rest, \( v(0) = 0 \), so \( C = 0 \). Thus, \[ v(t) = 0.05t^2 - 0.0033t^3 \]
04

Integrate velocity to find displacement

Integrate the velocity function to find the displacement as a function of time up to \( t = 10 \) seconds:\[ s(t) = \int v(t) \, dt = \int (0.05t^2 - 0.0033t^3) \, dt \]\[ s(t) = \left( \frac{0.05}{3}t^3 - \frac{0.0033}{4}t^4 \right) + C \]As the motion starts from rest, \( s(0) = 0 \), hence \( C = 0 \):\[ s(t) = \frac{0.05}{3}t^3 - \frac{0.0033}{4}t^4 \]
05

Calculate displacement at t = 10 seconds

Calculate \( s(10) \) using the displacement function:\[ s(10) = \frac{0.05}{3}(10)^3 - \frac{0.0033}{4}(10)^4 \]Simplifying these terms gives:\[ s(10) = \frac{0.05}{3} \cdot 1000 - \frac{0.0033}{4} \cdot 10000 \]\[ s(10) = 16.67 - 8.25 = 8.42 \, \text{meters} \]

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Key Concepts

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

Kinematics
Kinematics is a branch of mechanics that involves the study of motion without considering the forces that cause the motion. It primarily focuses on the trajectories and velocities of moving objects.
In this exercise, we examine motion starting from rest, which means initial velocity is zero. Key terms such as displacement, velocity, and acceleration define kinematic equations. These describe how objects move through space over time.
  • Displacement is the object's change in position, not necessarily covering the path taken.
  • Velocity describes the speed of an object in a specific direction. It is a vector quantity that changes as the object accelerates.
  • Acceleration is the change in velocity over time, indicating how quickly an object speeds up or slows down.
Understanding these concepts is essential, as they form the foundation for more complex topics such as dynamics.
Integration of Motion Equations
Once you understand kinematics, integrating motion equations is the next step. It's about determining velocity and displacement from acceleration.
Starting from the known force equation, \[ F(t) = 10t - t^2 \] we calculate acceleration from Newton's second law: \[ a(t) = \frac{F}{m} = 0.1t - 0.01t^2 \]
To find velocity, integrate the acceleration function over time. The process of integration accumulates total 'change' over time. Here's how it's done:
\[ v(t) = \int (0.1t - 0.01t^2) \, dt \] which gives us velocity in terms of time. The integration constant is determined by initial conditions.

Similarly, integrating the velocity function derives the displacement:\[ s(t) = \int v(t) \, dt \] By repeating this process, we derive the main motion equations used to analyze movement.
Displacement Calculation
Displacement calculation is the culmination of integration, representing how far an object has moved from its starting point.
This exercise solves for displacement when a force acts until the object stops accelerating. The velocity function:\[ v(t) = 0.05t^2 - 0.0033t^3 \]
when integrated, gives:\[ s(t) = \frac{0.05}{3}t^3 - \frac{0.0033}{4}t^4 \]
Calculate displacement at the critical time of 10 seconds, where force becomes zero:\[ s(10) = 16.67 - 8.25 = 8.42 \, \text{meters} \]
This result demonstrates the effectiveness of integrating kinematic equations for solving complex motion problems. The application of calculus in physics helps predict and describe motion accurately.

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Most popular questions from this chapter

A body with a mass \(m\) slides along the surface of a trihedral prism of mass \(M\), whose upper plane is inclined at an angle \(\alpha\) to the horizontal. The prism rests on a horizontal plane having a vertical wall at the rear edge of the prism to keep it at rest. The force exerted by the base of the prism on the horizontal plane is (a) \(M g\) (b) \(m g\) (c) \(M g+m g \sin ^{2} \alpha\) (d) \(M g+m g \cos ^{2} \alpha\)

\(\Lambda\) vertical force of magnitude \(F\) acts at the top of a string of mass \(m\) and length \(l\). If a body of mass \(M\) hangs at the bottom of the string. find the: (a) acceleration of the system \((M+m)\), (b) variation of tension in the string with \(x\).

A block of mass \(m\) connceted with a fixed point by a light inextensible string is kept on a smooth wedge of mass \(M\) and angle of inclination \(\theta\). If the string is parallel to the inclinc at the time of its release, find the (a) lension in the string, (b) accelerations of \(M\) and \(m\), (c) contact forces belween \(M\) and \(m\), (d) reaction force oflered by ground on the wedge.

Which of the following forces are electromagnetic in nature? (a) Reaction on a body (b) Tension in a spring (c) Weight of a body (d) Elastic force in a string

Statement-1 : In a frame of reference attached to a particle itself, the acceleration of the particle is zero. Statement-2: The frame of reference which moves with constant velocity w.r.t. another frame of reference is an inertial frame of reference.
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