How Do You Evaluate the Extension Value Under a 6N Load Using F = kx?

[r-soy]

evaluate the value of extension ( x) which elongate under the action of load (w) of 6 N ? [Use your results obtained from rest 1 ]



please help me How i can solve this question because I don't understand it clearly

 

[r-soy]

plese I want your help

 

[r-soy]

any one here ??

 

[Mark44]

What is the question? What do the three columns on the right mean? Why do you have the same values for Increasing and Decreasing?

 

[r-soy]

beacuse when we mesure we got same value to Increasing and Decreasing

 

[Mark44]

I can see that, but I have no idea what you're trying to do here. What do Increasing and Decreasing mean in your table?

 

[r-soy]

this experiment of Speing balance

when we increasind ( Load ) the space we are got in decreasing is same .

we start by 0.1 N then 0.2 then >>>> to 0.5

 

[r-soy]

where are you ?

 

[Mark44]

OK, for the first row in your table, you have Load = .1 N. Did the spring length change (increase) by 29 mm? If the spring got longer, it can't have also gotten shorter, so I don't understand the purpose of the Decreasing column or the Average column.

 

[r-soy]

Look Dear

Experience as follows :

First, we put Spring on the board and a paper under spring to measure the distances between each point from the beginning to the end point of extension for each weight

Then we increased laod to ( 0.2 n ) the measurement was between the starting point until end point of extension is 51 mm (( same steps for all load ))

Then we've decrease the weights ( Every time one ) from end ( 0.5 ) to (0,1 ) we got same measurements

then We calculated the average by calculat increases and decreases and divide by 2




For information, there is a Positive relationship between weight and extension
plese help me

 

[r-soy]

Dr. Mark44 where are you ?

 

[Mark44]

It looks to me like the spring will extend about 27 mm per .1 N, or about 2.7 cm per N. How much will the spring extend with a weight of 6.0 N?

 

[r-soy]

Is this ok help me

k = mg/x

0..04/1 = 6 / X

x = 6.6X10^-6 cm

plese help me

 

[Mark44]

Is this ok help me

k = mg/x

0..04/1 = 6 / X

Where did the equation above come from, particularly "0..04/1"?

x = 6.6X10^-6 cm

No, this is way off.

 

[r-soy]

x = w/k

but from where we got the value of k ?

 

[Mark44]

You have the force (mg - Newtons), and you have the displacement (\Delta x).

 

[r-soy]

x = w/k =
6/k from where we got the value of k ? here we try to find x

 

[Mark44]

The purpose of your experiment was to measure values of F (the weight) and x.

 

[r-soy]

plese i can't understand help me

 

[r-soy]

plese help me

 

[Mark44]

The formula is F = kx.
The table you made has measured values of F and x, so solve for k in the equation F = kx.