# Homework Help: Very easy question: why can't I use all 3 of these constant-acceleration formulas?

1. Sep 28, 2012

### PhyIsOhSoHard

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Newton's second law:
$\Sigma F=ma$
$v^2_x=v^2_{0x}+2a_x(x-x_0)$
$v_x=v_{0x}+a_xt$

3. The attempt at a solution
$\Sigma F=80N$
$v_x=\frac{11m}{5s}=2.2m/s$
$v_{0x}=0m/s$
$x=x_0+v_{0x}t+1/2a_xt^2$

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:
$(2.2m/s)^2=0+2a_x(11m)$
Which gives: $a_x=\frac{(2.2m/s)^2}{2\cdot 11m}=0.22m/s^2$

But if I use the 2nd formula:
$2.2m/s=0+a_x\cdot 5s$
Which gives: $a_x=\frac{2.2m/s}{5s}=0.44m/s^2$

And if I use the 3rd formula:
$11m=0t+1/2a_x\cdot (5s)^2$
Which gives: $a_x=\frac{11m}{1/2\cdot (5s)^2}=0.88m/s^2$

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

Last edited: Sep 28, 2012
2. Sep 28, 2012

### hqjb

Re: Very easy question: why can't I use both of these constant-acceleration formulas?

Your $V_x$ is average velocity and not final velocity.

3. Sep 28, 2012

### PhyIsOhSoHard

Re: Very easy question: why can't I use both of these constant-acceleration formulas?

I've updated my 1st post with another equation. According to my book, it says the 3rd equation is the correct with the acceleration being 0.88 m/s^2

But that is the average velocity as well?

4. Sep 28, 2012

### SammyS

Staff Emeritus
Re: Very easy question: why can't I use both of these constant-acceleration formulas?

$v_x$ is the final velocity of the block,

whereas $22 \text{m/s}$ is the average velocity of the block.

For constant acceleration, the average velocity is given by: $\displaystyle v_\text{Average}=\frac{v_x+{v_0}_x}{2}\ .$

5. Sep 28, 2012

### hqjb

Re: Very easy question: why can't I use both of these constant-acceleration formulas?

The third equation is $S = V_{0x}t + \frac{1}{2}at^2$
where S = distance, V0x = initial velocity a = constant acc. t = time, no final velocity needed.

Acceleration is constant therefore, velocity is changing throughout, the actual final velocity is not given in the question and can be calculated using v = u + at to be 4.4m/s