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## Homework Statement

A dockworker applies a constant horizontal force of 80 N to a block of ice on a smooth horizontal floor. The fritional force is negligible. The block starts from rest and moves 11 meter in 5 seconds.

What is the mass of the block of ice?

## Homework Equations

Newton's second law:

[itex]\Sigma F=ma[/itex]

[itex]v^2_x=v^2_{0x}+2a_x(x-x_0)[/itex]

[itex]v_x=v_{0x}+a_xt[/itex]

## The Attempt at a Solution

[itex]\Sigma F=80N[/itex]

[itex]v_x=\frac{11m}{5s}=2.2m/s[/itex]

[itex]v_{0x}=0m/s[/itex]

[itex]x=x_0+v_{0x}t+1/2a_xt^2[/itex]

Now in order to use Newton's second law, I need to find the acceleration.

My question is, why can't I use all 3 of the formulas?

When I use the first formula:

[itex](2.2m/s)^2=0+2a_x(11m)[/itex]

Which gives: [itex]a_x=\frac{(2.2m/s)^2}{2\cdot 11m}=0.22m/s^2[/itex]

But if I use the 2nd formula:

[itex]2.2m/s=0+a_x\cdot 5s[/itex]

Which gives: [itex]a_x=\frac{2.2m/s}{5s}=0.44m/s^2[/itex]

And if I use the 3rd formula:

[itex]11m=0t+1/2a_x\cdot (5s)^2[/itex]

Which gives: [itex]a_x=\frac{11m}{1/2\cdot (5s)^2}=0.88m/s^2[/itex]

What's the difference between the three formulas? Why do I get different accelerations?

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