Least amount of time for motion

[aaphys]

Homework Statement

: [/B]

The coefficient of static friction is 0.65 between two blocks. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force

(a pull applied to the top block) causes both blocks to cross a distance of 5.0 m, starting from rest. What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

Let:
topBlock = m1
lowerBlock = m2

Given:
m1 = 4.0 kg
m2 = 3.0 kg
μs = 0.65
μk = 0.2
Δd = 5.0 m

Homework Equations

:[/B]
∑F = ma
Δd = (1/2)at^2

The Attempt at a Solution

:
[/B]
Starting with Δd = (0.5)at^2

a = [2Δd]/[t^2]

Now using ∑F = ma

ΣF = [m1+m2]*[a] = [m1+m2]*[2Δd]/[t^2]
ΣF = [2Δd*(m1+m2)]/t^2
t^2 = [2Δd*(m1+m2)]/[ΣF]

t^2 = [2*5*(4+3)]/[ΣF]
t^2 = [70]/[ΣF]

To find ΣF I said that the force applied must be one that is

m1gμs > F > (m1+m2)gμk → (4)*(9.8)*(0.65) > F > (4 + 3)*(9.8)*(0.2) → 25.48N > F > 13.72N

Then I use 25.48N and

t^2 = 70/25.48
t = √[70/25.48]
t= 1.657 s

And it's wrong :(

 

[andrewkirk]

Your calculation looks correct to me. Maybe somebody else can spot a problem we missed, or maybe the answer in the book is wrong.

 

[OmCheeto]

Your calculation looks correct to me. Maybe somebody else can spot a problem we missed, or maybe the answer in the book is wrong.

I also got the same answer.