# Two forces on a particle

1.) A particle experiences two forces. One constant force is given as F = F_0 x(^) (vector F = F(initial) * x(unit vector)) and the other force is a drag force that is proportional to the velocity with proportionality constant ÃŸ.

Write out Newton?s 2nd Law for this particle and solve the differential equation for the particle?s position along the x-axis as a function of time. Show that if the particle starts from rest, the velocity will reach a limiting value of F_0/ÃŸ.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

The total force on the particle is:

(1.1)

Then, Newton's second law states that:

(1.2)

Where M is the mass, and a is the acceleration vector.

Since we are dealing with one dimensional motion (the particle starts from rest, so the subsequent motion will always be along the x axis), we can drop the vector notations ad write:

(1.3)

But the acceleration is the first derivative of the velocity with respect to time:

(1.4)

Then, the equation becomes a first order differential equation:

(1.5)

Where

(1.6)

Equation (1.5) is a separable equation:

(1.7)

Integrating both sides yields:

(1.8)

Where C is an arbitrary constant of integration.

Form initial conditions:

(1.9)

Thus:

(1.10)

We now use (1.6) to write:

(1.11)

We see that as the exponent approaches zero, hence:

(1.12)

The speed is the first derivative of the displacement with respect to time.

Hence, to find the position all we have to do is to integrate (1.11) with respect to time:

(1.13)

Where C is an arbitrary constant of integration.

If at t=0 the particle is in some arbitrary origin, then:

(1.14)

So the position of the particle at any given time is:

(1.15)

Or:

(1.16)

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