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Boxes are moved on a converyor blet from where they are filled to the packing station 10m away. The belt is initially stationary and must finish with zero speed. The most rapid transit is accomplished if the belt accerlates for half the distance, then decelerates for the final half of the trip. If the coefficent of static friction between a box and the belt is .60, what is the mininum transit time for each box?

I have no idea what it's talking about or how many boxes there are so I just assumed one...

acceleration first half of trip
i got 5.880 s^-2 m

time for first half
i got 4.124 s

velocity at end of first half
i got 7.668 s^-1 m

acceleration for second half
i got - 6.380 s^-2 m

time for second half
i got 1.202 s

total time
i got 5.3 s

This problem was kind of more involved and I got 5.3 s and listed all of the other variables I solved for in order to get that answer... Does it look right???

• physics -

"acceleration first half of trip
i got 5.880 s^-2 m "
OK, as it equals 0.6g, the maximum acceleration without the box slipping from the belt.

I do not know how you got the time for the first half, my formula is
S = ut + (1/2)at²
for initial velocity u=0, a=5.88, and distance S=10/2=5m
t=√(2S/a)=10/5.886=1.303 s.
So the minimum time is twice this value, 2*1.303=2.606 s., because the deceleration takes the same time as the acceleration.
The boxes may slip a little forward as the acceleration changes to deceleration, but that will only make the box arrive a fraction of a second earlier.

• physics -

Acceleration for first half of trip...
apply newtons second law...

net force = m a = - F fr = - (Mu s Fn)
were Mu s is coefficent of static friction

take the inverse of m onto both sides

a = m^-1 -(Mu s Fn)

apply Newtons second law in the y direction

net force (y direction) = ma = FN - Fg = 0

0 sense it's not moving in this direction and FN is the normal force

FN = Fg = mg

plug this value of FN into the equation for Newtons Second Law in x direction which was rearanged for a...

a = m^-1 -(Mu s Fn)=m^-1 -(Mu s mg)

cross out the masses

a = - (Mu s g )= -(.60 (-9.80 s^-2 m))

two negatives cross out giving positive

a = 5.880 s^-2 m

X = Xo + Vo t + 2^-1 a t^2

Xo is zero and so is Vo

X = 2^-1 a t^2

solve for t

t = (a^-1 2 x)^(2^-1)

t = (5.880 s^-2 m (2) 50 m)^(2^-1)

t = 4.124 s

velocity as half

v = (2a (X - Xo) + Vo^2)^(2^-1)

Xo is zero and so is Vo^2

v = (2a X)^(2^-1)

v = (2 (5.880 s^-2 m) 5.0m)^(2^-1)

v = 7.668 s^-1 m

acceleration second half

a x = (V^2 - Vo^2) + Xo

solved for a were V^2 is zero

a = x^-1(-Vo^2 + Xo)
a = (10 m)^-1 (-(7.668 s^-1 m)^2 + 5.0 m)

a = -6.380 s^-2 m

time for second half

a = t^-1 (V - Vo)

t = a^-1 - Vo

t = (- 6.380 s^-1 m)^-1 -(7.668 s^-1 m)

t = 1.202 s

total t = 1.202 s + 4.124 s

t = 5.3 s

how do you incorporate that theres is static friction i thought I did

• physics -

The static friction is incorporated, as mathmate indicated, by assumeing the max acceleration and deaccelearation was .6 g.

Fmax=fn*mu= mg*.6
Fmax=m(accelerationmax)=mg*.6
accelerationmax=.6g

• physics -

I did that

• physics -

if you do this

t=ã(2S/a)=10/5.886=1.303 s.

your doing the full distance 10 m instead of 5 m right???

so why would you double it

• physics -

I believe there is a discrepancy of the data.

According to the posted question,
"Boxes are moved on a converyor blet from where they are filled to the packing station 10m away..."
with which 1.304 seconds for half of the trip was obtained, using the formula
t=sqrt(2*S/a) where S=half of total distance and a=acceleration.

Your calculation probably assumes a half-distance of 50m to get
"t = (5.880 s^-2 m (2) 50 m)^(2^-1)
t = 4.124 s "

The second half of the trip should take exactly the same time as the first half, whether the distance is 100 or 10 m.

• physics -

the doubling comes from:

S=ut+(1/2)at²
when u=0,
S=(1/2)at²
t²=2*s/a
t=sqrt(2*s/a)

• physics -

oh wow...
Thanks!!!

• physics -

You're welcome!

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