Newtonian mechanics on a particle

[jlmac2001]

Say that a particle of mass m slides down an inclined plane under the influence of gravity. If he motion is resisted by a force f=kmv^2, how canyou show that the time requied to move a distanc d after starting form rest is t= cosh^-1(e^kd)/(sqrt kg sin theta) where theta is the angle f inclination o the plane.

 

[HallsofIvy]

When will people learn that it is impossible to give ideas about how to solve some of these problems without knowing what they have to "work with"! That's one good reason to SHOW WHAT YOU HAVE DONE SO FAR and what you do understand about the problem. I would set this up as a differential equation and then solve that equation but that may not be the way you would need to do it.

The first thing you would do is draw a diagram! You should then see from the diagram that the component of gravitational force (weight) down the slope is -mg cos(θ). Since there is resisting force kmv², The total force, taking "+" up the inclined plane is kmv²- mg. Since "force= mass times acceleration": mdv/dt= kmv²- mg, a differential equation.
That's relatively easy to solve (If you have studied differential equations!) but you don't really have to do that. You asked "how can you show that the time requied to move a distanc d after starting form rest is t= cosh^-1(e^kd)/(sqrt kg sin theta)". Since you are given tis you can "turn it around" by solving for d (that's relatively easy since the obvious inverse of cosh^-1() is just cosh(). After you have d as a function of t, differentiate to get v as a function of t. Now plug that into the equation and show that it satisfies it.

 

[M98Ranger]

To show the time required for the particle to move a distance d after starting from rest, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F=ma). In this case, the net force acting on the particle is the sum of the gravitational force and the resisting force, which is given by f=kmv^2.

Since the particle is sliding down an inclined plane, we can break down the gravitational force into its components parallel and perpendicular to the plane. The parallel component is given by mgsinθ, where θ is the angle of inclination of the plane, and the perpendicular component is mgcosθ.

Using Newton's second law, we can set up the following equation:

ma = mgsinθ - kmv^2

We can rearrange this equation to solve for acceleration (a):

a = (g sinθ - kv^2)/m

Next, we can use the kinematic equation for displacement (d=vt+1/2at^2) to find the time required for the particle to move a distance d. Since the particle starts from rest, the initial velocity (v0) is 0. This gives us the following equation:

d = 1/2at^2

Substituting our expression for acceleration (a) into this equation, we get:

d = 1/2(gsinθ - kv^2)t^2/m

Solving for time (t), we get:

t = √(2md/(gsinθ - kv^2))

Now, we need to find an expression for the final velocity (v) of the particle. Since the particle is starting from rest, we can use the kinematic equation for final velocity (v=v0+at). Again, since v0=0, we get:

v = at = (gsinθ - kv^2)t/m

We can substitute this expression for velocity (v) into our equation for time (t) to get:

t = √(2md/(gsinθ - k(gsinθ - kv^2)^2/m^2))

Simplifying this expression, we get:

t = √(2md/(gsinθ - kgsin^2θ/m))

Using the identity cosh^-1(x) = ln(x+√(x^2-1