Friction and Distance and Time

[Nbwx]

Homework Statement

The coefficient of static friction is 0.673 between the two blocks shown. The coefficient of kinetic friction between the lower block and the floor is 0.113. Force F causes both blocks to cross a distance of 7.20m, starting from rest. What is the least amount of time in which the motion can be completed without the top block sliding on the lower block, if the mass of the lower block is 1.90kg and the mass of the upper block is 2.54kg?

Homework Equations

Force Friction = μ*Force Normal
Δd= vt +0.5aΔt²

The Attempt at a Solution

From what I know is that the force static friction is the total amount of force allowed without allowing it to slip.
Force Friction = μ*Force Normal
Force Friction = 0.673*(mass of top block*gravity)
Force Friction = 0.673*(2.54*9.8)
Force Friction = 16.75 N

Now F=ma
16.75/(mass of top+bottom)=a
16.75/4.44=a
3.77 m/s^2 = a

Now Δd= vt +0.5aΔt²
7.2= 0*t +0.5(3.77)Δt²
Solve for t...1.95 seconds...wrong...

What Am I doing wrong?

 

[AxM=Fam]

I believe all your formulas and calculations look correct but I would think about why it would be necessary to add both masses when looking for the acceleration. It is only the maximum acceleration of the top block that is needed.

 

[gneill]

Homework Statement

The coefficient of static friction is 0.673 between the two blocks shown. The coefficient of kinetic friction between the lower block and the floor is 0.113. Force F causes both blocks to cross a distance of 7.20m, starting from rest. What is the least amount of time in which the motion can be completed without the top block sliding on the lower block, if the mass of the lower block is 1.90kg and the mass of the upper block is 2.54kg?

Homework Equations

Force Friction = μ*Force Normal
Δd= vt +0.5aΔt²

The Attempt at a Solution

From what I know is that the force static friction is the total amount of force allowed without allowing it to slip.
Force Friction = μ*Force Normal
Force Friction = 0.673*(mass of top block*gravity)
Force Friction = 0.673*(2.54*9.8)
Force Friction = 16.75 N

Now F=ma
16.75/(mass of top+bottom)=a
16.75/4.44=a
3.77 m/s^2 = a

Now Δd= vt +0.5aΔt²
7.2= 0*t +0.5(3.77)Δt²
Solve for t...1.95 seconds...wrong...

What Am I doing wrong?

Hi Nbwx, Welcome to Physics Forums.

Is the external force being applied to the top or bottom block (or is it your choice to choose)?

 

[Nbwx]

I believe the force is being applied to the top block.

 

[gneill]

I believe the force is being applied to the top block.

In that case the maximum horizontal force that can be transmitted to the bottom block via the top block is what? What force impedes the bottom block's motion? Can you draw a FBD for the bottom block including these forces?