Simple Force and Motion Problem - Explain the explanation please.

[tenbee]

Homework Statement

Half-Passage: In the packing industry, processing packages via conveyor belts is immensely practical and vital for operations. Maintaining packages in a neat and file line is no easy task, considering that many conveyor belts are non-linear and require the traversal of hills and valleys. As a general rule of thumb, boxes are kept no less than 5m apart so that collisions do not occur and skew the entire process. In the following diagram, box A is at the very top of the conveyor belt and box B is at the very bottom. The height of the conveyor belt system is h and the slops of the hill is r.

http://www.mcatquestionaday.com/pictures/080925.gif

Question 2: Assume that the conveyor belt of the incline is distinct and moves separately from the conveyor belt on which B rests. If the conveyor belt underneath A breaks, how long will it take for box A to hit box B assuming that box B remains stationary and the coefficient of kinetic friction of the conveyor belt is 0.4? Also assume that the mass of box A is m. For simplicity, assume that the distance down the ramp is d.

Homework Equations

The Attempt at a Solution

answer: {2d/[gsin(r) - 0.4gcos(r)]}1/2

Okay... I understand how to solve this equation, but I don't understand why it's '2d' instead of 'd'. A little help please : )

To find time...
a = v/t and v= d/t, so a = (d/t)/t
F = ma --> a = F/m
F_fr = µ_k*F_norm

So for force --> (mgsinθ - mgcosθ)/m, then cancel the m --> a = (gsinθ - gcosθ)

(d/t)/t = (gsinθ - gcosθ) --> (d/t) = t(gsinθ - gcosθ) --> d = t*t(gsinθ - gcosθ --> d/(gsinθ - gcosθ) = t² --> [d/(gsinθ - gcosθ)]^1/2 = t

Where do they get 2d from?!?

 

[khemist]

Try setting a = 1/2 t^2, taken from x(t) = 1/2t^2 + vt + t_0

 

[tenbee]

Try setting a = 1/2 t^2, taken from x(t) = 1/2t^2 + vt + t_0

Ahhh, I see - yes that works. Thank you!

When should I use x = x₀ + v₀t + 1/2at² versus a = (d/t)/t?

 

[khemist]

x(t) is a position function. It implies constant acceleration and one can use it to determine the time it takes to get from A to B .

I would only use a = d/t^2 when you know the distance interval and time interval. It is really a = (d_2-d_1)/(t_2 - t_1)^2, which gives you an average acceleration.

Someone else might be able to answer that question a little better, I am not 100% on when to use it.

 

[jaumzaum]

Actually the problem is wrong. If A is moving, the conveyor has a velocity v, and when it stops, it makes a get an disacelleration or an acelleration (only if static friction coefficient makes A don't slide in the conveyor, but kinetic does). This way A has a initial velocity v that increases or decreases by time, so the initial velocity is needed. But as the answer says, there is no v, so, assuming that v is TOO short and can be forgotten, we have:

We know the position/time function in the MUV:

\Delta S = Vot - (1/2)at²
Vo=null
\Delta S = (1/2)at²
t=\sqrt{2 \Delta S/a}

But a is the acceleration of gravity in the axe of the conveyor minus the disacxceleration of friction

A = g.sin\alpha - A_{friction}

Fa = N.u = m.g.cos\alpha.u -> Fa = 0.4mgcos\alpha -> A_{ friction} = 0.4gcos\alpha
A = g.sin\alpha - 0.4gcos\alpha
t = \sqrt{ 2\Delta S/g.sin\alpha - 0.4gcos\alpha}

Replace to get answer