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

  • #1

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?
 
Last edited:

Answers and Replies

  • #2
40
0


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]

Now in order to use Newton's second law, I need to find the acceleration.
My question is, why can't I use both 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 other 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]

What's the difference between the two formulas? Why do I get different accelerations?
Your [itex]V_x[/itex] is average velocity and not final velocity.
 
  • #3


Your [itex]V_x[/itex] is average velocity and not final velocity.
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
SammyS
Staff Emeritus
Science Advisor
Homework Helper
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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 frictional 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]

Now in order to use Newton's second law, I need to find the acceleration.
My question is, why can't I use both 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 other 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]

What's the difference between the two formulas? Why do I get different accelerations?
[itex]v_x[/itex] is the final velocity of the block,

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

For constant acceleration, the average velocity is given by: [itex]\displaystyle v_\text{Average}=\frac{v_x+{v_0}_x}{2}\ .[/itex]
 
  • #5
40
0


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?
The third equation is [itex]S = V_{0x}t + \frac{1}{2}at^2[/itex]
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
 

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