# Block on incline with air resistance, no friction

• inferno298
In summary, the conversation discusses a textbook of mass 1.24 kg sliding down a frictionless inclined plane with an angle of 30 degrees. The problem includes a drag force due to air resistance that is proportional to the speed squared, with a constant of 0.86 m^-1. The question asks for the time it takes for the textbook to slide a distance of 1.65 m down the plane. The conversation includes attempts at solving the problem, but there is a discrepancy with the final time calculation due to a missing negative sign and potential incorrect data.

## Homework Statement

A textbook of mass m = 1.24 kg starts at rest on a
frictionless inclined plane (angle  = 30◦). Although there is
no friction, suppose there is a drag force (due to air resistance)
acting on the book which is proportional to the speed squared
and is described by the equation F = kmv2, where k = 0.86
m−1. How much time does it take for the textbook to slide a
distance d = 1.65 m down the plane? (Hint: This one is tricky,
you will need to solve the integral by hand using a hyperbolic
trig substitution.)

## Homework Equations

Just F=m*a
drag force = k*m*v^2

## The Attempt at a Solution

So the mass keeps canceling on me whenever I run the problem, don't know if I actually need it for this one. I changed my coordinate system so positive x was parallel to the incline and positive y was perpendicular to it.

Forces are gravity broken in componets now, normal force only acting in positive y, canceled out by gravities y component. drag force in -x direction and +x component of gravity.
Vf= final velocity
y- direction F=ma
F= 0, its moving down the incline

x - direction F= ma
m*g*cos(theata)-k*m*v^2 = m dv/dt
solve it for time with given conditions and I got
T = (1/(Sqrt[g*sin(theata)*k])*ArcTanh[Sqrt[k/(g*sin(theata))]*Vf]

Switch the beginning equation to m*g*cos(theata)-k*m*v^2 = m*v* dv/dx and solve for the distance given the distance yields:
x = (1/2*k)*Ln((g*sin(theata)-k*Vf^2)/(g*sin(theata)))
Solve that equation for Vf yields:
Vf = Sqrt[(g*sin(theata)/k)*(1-e^(2*k*x))/k)
Sub into T equation and get:
T = (1/(Sqrt[g*sin(theata)*k])*ArcTanh(1-e^(2*k*x))

I keep getting very small times less than one and complex numbers which I feel are wrong, well its telling me they are wrong. I might have set it up wrong when defining my coordinate system or something. if someone can help me that would be awesome, even if its on a regular defined coordinate axes.

It's an old post so I'm not sure if you still want advice, but I think the issue comes with your second diff eq you set up to solve for the distance.

You can save a lot of work and simply turn your first diff eq into a function of distance right off the bat by using derivatives of ##x(t)## in place of ##v## and ##a##. This can be integrated without trouble given 2 initial conditions on velocity and position. From there, you should have a function ##x(t)## that you can use to plug in your numbers and numerically invert. It seemed to give a reasonable real value for time when I worked it out.

inferno298 said:
Switch the beginning equation to m*g*cos(theata)-k*m*v^2 = m*v* dv/dx and solve for the distance given the distance yields:
x = (1/2*k)*Ln((g*sin(theata)-k*Vf^2)/(g*sin(theata)))