the problem is with b. and c. but I would like verification on a if possible, thanks!(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

A cabin lays on a smooth track (so friction can be disregarded)

the wind exerts a pushing force [itex]F=\gamma u[/itex] u is the relative velocity between

the cabin and the wind. the wind's velocity is [itex]v_0[/itex]

a. what is the maximum velocity?

b. how much time will it take to the cabin to reach 86% of this velocity?

c. sand is spread on the track and the friction coefficient is [itex]\mu_k[/itex]

what is the maximum velocity in this situation?

2. Relevant equations

[itex]F=ma[/itex]

[itex]a=\frac{dv}{dt}[/itex]

3. The attempt at a solution

a. let's assume the velocity of the cabin is [itex]v_c[/itex] then:

[itex]u=v_0 - v_c[/itex]

the maximum velocity will be reached when there is no more acceleration.

[itex]ma=\gamma u = 0[/itex]

so [itex] u=0 , v_c=v_0[/itex]

so the maximum velocity is [itex]v_0[/itex]

b.

[itex]F=ma=m\frac{du}{dt} , m\frac{du}{dt}=\gamma u[/itex]

multiplying by dt dividing by m and u we get:

[itex]\frac{1}{u}du=\frac{\gamma}{m}dt[/itex]

integeral on both sides from final to initial:

[itex]u(t)=v_0 - v_c (t)[/itex]

[itex]\int_{v_0}^{u(t)}\frac{1}{u}du=\int_{0}^{t}\frac{\gamma}{m}dt[/itex]

this gives:

[itex]ln\frac{u(t)}{v_0}=\frac{\gamma}{m}t[/itex]

[itex]\frac{v_0 - v_c (t)}{v_0}=e^{\frac{\gamma}{m}t}[/itex]

so the velocity(time):

[itex]v_c (t)=v_0 (1- e^{\frac{\gamma}{m}t})[/itex]

this can't be right - more logical is [itex]v_c (t)=v_0 (1- e^{-\frac{\gamma}{m}t})[/itex]

but I don't see where my error is.

let's assume I have the right velocity, we need to find when it's 86% so

[itex]v_c (t)= 0.86v_0[/itex]

[itex]e^{-\frac{\gamma}{m}t}=0.14[/itex]

[itex]t=-\frac{mln0.14}{\gamma}[/itex]

c. now [itex]F= \gamma u -mg\mu_k[/itex]

but I'm not sure how I should proceed

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# Homework Help: Movement caused by wind (changing force)

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