Particle of mass ‘m’ sits at rest at x=0 at time t=0 a force given by F= F_oe^-t/T is applied in the positive direction (+ x direction) where f_o and T are constant when t=T the force is removed out this instant when the force is removed find

1. The speed of particle?

The position of the particle

To solve this problem, we need to analyze the motion of the particle under the influence of the given force. The force applied is a time-dependent force, which means we need to integrate it to find the particle's velocity and position over time. Let's break this down step by step.

Understanding the Force

The force acting on the particle is given by:

F(t) = F₀ e^(-t/T)

Here, F₀ is the initial force, T is a constant time parameter, and e is the base of the natural logarithm. This force decreases exponentially over time.

Finding Acceleration

According to Newton's second law, the acceleration a of the particle can be expressed as:

a(t) = F(t) / m

Substituting the expression for force, we get:

a(t) = (F₀ / m) e^(-t/T)

Calculating Velocity

To find the velocity of the particle, we need to integrate the acceleration with respect to time:

v(t) = ∫ a(t) dt = ∫ (F₀ / m) e^(-t/T) dt

The integral of e^(-t/T) is:

-T e^(-t/T)

Thus, we have:

v(t) = - (F₀ T / m) e^(-t/T) + C

Since the particle starts from rest at t = 0, we can find the constant C:

v(0) = 0 = - (F₀ T / m) + C

Therefore, C = (F₀ T / m) and the velocity equation becomes:

v(t) = (F₀ T / m) (1 - e^(-t/T))

Evaluating Velocity at t = T

Now, we need to find the velocity at the moment the force is removed, which is at t = T:

v(T) = (F₀ T / m) (1 - e^(-1))

Using the approximate value of e^(-1) ≈ 0.3679, we can simplify this to:

v(T) ≈ (F₀ T / m) (1 - 0.3679) ≈ (F₀ T / m) (0.6321)

Finding Position

Next, we need to determine the position of the particle. The position x(t) can be found by integrating the velocity:

x(t) = ∫ v(t) dt

Substituting the expression for velocity, we get:

x(t) = ∫ [(F₀ T / m) (1 - e^(-t/T))] dt

This integral can be separated into two parts:

• ∫ (F₀ T / m) dt = (F₀ T / m) t
• ∫ (F₀ T / m) e^(-t/T) dt = - (F₀ T² / m) e^(-t/T)

Combining these results, we find:

x(t) = (F₀ T / m) t - (F₀ T² / m) e^(-t/T) + C'

At t = 0, the particle is at rest at x = 0, so we can determine the constant C'. Thus, C' = 0, leading to:

x(t) = (F₀ T / m) t - (F₀ T² / m) e^(-t/T)

Evaluating Position at t = T

Now, substituting t = T into the position equation gives:

x(T) = (F₀ T² / m) - (F₀ T² / m) e^(-1)

Using the approximation for e^(-1), we have:

x(T) ≈ (F₀ T² / m) (1 - 0.3679) ≈ (F₀ T² / m) (0.6321)

Final Results

In summary, when the force is removed at t = T, the speed of the particle is:

v(T) ≈ (F₀ T / m) (0.6321)

And the position of the particle at that instant is:

x(T) ≈ (F₀ T² / m) (0.6321)

This analysis shows how the particle's motion is influenced by the time-dependent force, leading to both a specific speed and position at the moment the force is removed.