An object of mass m is at rest in equilibrium at the origin. At t=0 a new force F(t) is applied

An object of mass m is at rest in equilibrium at the origin....

Key Concepts

Newton's Second Law

This principle states that the net force acting on an object is equal to the mass times its acceleration (F = ma). It forms the basis for deriving the equations of motion by relating applied forces to the resulting acceleration, which then must be integrated to obtain velocity and position.

Ordinary Differential Equations (ODEs)

When forces vary with time or position, the equations of motion become differential equations. In this case, the acceleration components are given as functions of time and position, leading to second-order ODEs that describe how the velocity and position of the object evolve over time.

Integration of Equations of Motion

Once differential equations are formulated using Newton's second law, they must be integrated to obtain the velocity and position functions. This process involves calculating indefinite integrals while applying the appropriate initial conditions to determine the constants of integration.

Initial Conditions

The initial state of the system, such as the object being at rest at the origin, provides the necessary boundary conditions. These conditions are vital to uniquely determine the integration constants and fully specify the solution to the differential equations.

Coupled Differential Equations

In situations where the force in one direction depends on the position in another (as with the dependency of F_x on y), the resulting equations are coupled. Solving coupled differential equations often involves solving one equation first and then substituting its solution into the other to decouple the system for integration.

Transcript

00:01 Okay, so today we're going to talk about newton's second law of motion.

00:06 So remember that newton's second law says that the net force in a body is equal to its mass times its acceleration.

00:22 Okay? now, in the exercise, we're going to face the following setup.

00:27 So we have a system.

00:30 So this system is an object of mass m.

00:36 And we know that when t is equal to zero, we are going to exert an external force on its mass, in its mass, sorry.

00:53 And the expression for this external force is the following.

00:58 So it is a time -dependent force that is equal to, k1 plus k2y in the x direction plus k3 times t in the y direction.

01:26 This is the expression for the force.

01:30 Now, the exercise wants us to give in this force, what is going to be the position and the development of the object as a function of time, and also knowing that k1, k2, and k3 are constants.

02:00 Okay? okay, so we can use newton's second law for the components of the motion.

02:10 So we have in the x direction, the net force acting on the body will be k1, k1 plus k2 times y.

02:23 This is going to be equal to the mass times the acceleration in the x direction.

02:31 Okay, we can isolate the acceleration.

02:36 This is k1 plus k2y divided by m.

02:44 Now, recalling that the acceleration is the time derivative of the velocity.

02:52 So we have that the x component of the velocity.

02:59 Sorry, this is equal to dvx d t.

03:04 Okay.

03:05 Now, we can pass the time derivative to the other side of the equation.

03:11 So we have the infinitesimal velocity, divx, is going to be equal to k1 plus k2 times y.

03:24 Everything divided by m times d t okay we can integrate both sides so this side we integrate in vx so we are going to start from t equals zero and the exercise tells us that when t is equal to zero the initial velocity is also equal to zero and we are going to calculate for a time t so vxt.

03:58 Okay? so doing this integral, we have that vxt is going to be equal to k1 plus k2y divided by m times t.

04:16 So let's keep this expression in a second.

04:22 So i'm going to call it expression 1.

04:26 Now this is from newton's second law in the x direction.

04:30 Now in the y direction, i'm going to have the mass times the acceleration in the way direction is going to be equal to k3 times t.

04:46 Okay? once again, we have the d, vy, dt is going to be equal to k3 times t divided by m, passing dt to dt to the other side and integrating both sides from t equal to zero to t so the initial velocity to a final velocity that is time dependent we're going to have that the y component of the velocity is going to be k3 times t squared over 2m we can sum expressions 1 and 2 not some actually we can write the vector v as the sum of the components of its x and y components.

05:56 So vt is going to be equal to k1 plus k2y divided by m times t in the x direction, plus k3 t squared over 2m in the y direction.

06:26 Now, we could call the velocity a day and move and not touch into this expression anymore, but notice that we have this dependence of the position in the y direction here.

06:47 And the exercise asks us explicitly to find this as a function only of the two.

06:57 Time.

06:59 So since the only constants that we have here are k1, k2, k2, k2, k3, and m, that is the mass of the object, then it is going to be interesting to go back to this expression to substitute this term by the y equation as a function of time.

07:25 Okay...

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