- #1

aaphys

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## 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

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 :(