# General Physic I

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### 2 Answers

From the graph we see that for a 1N force there is an acceleration of `40 "m"/"sec"^2` .

We use the Newtonian Law F=ma . Force is measured in Newtons where `1N=1"kgm"/"sec"^2`

Substituting the given values into F=ma we get:

`1N=(x"kg")(40"m"/"sec"^2)` The units divide out (cancel) so we get:

`x=1/40` .

So the objects mass is `1/40 "kg"`

`1/40"kg"=1000/40"g"=25"g"`

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The mass is 25 grams

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First a bit of theory. For an elastic band that is stretched by a force `F` the elongation of the band `x` is proportional to the applied force (as long as the band is in the elastic region).

`F =k*x`

If we couple in parallel two elastic bands having the elastic constants `k_1 and k_2` the equivalent elastic constant obtained is

`1/k_(eq) =1/k_1 +1/k_2`

For n identical bands coupled in parallel we have `k_(eq) =k/n`

If we couple in series two elastic bands having elastic constants `k_1` and `k_2` the equivalent elastic constant obtained is

`k_(eq) =k_1+k_2`

The experiment should be as follows.

First take one band. Fix one end of it and at the other end stretch the band with with increasing forces of 1 N, 2 N, 3 N,..,10 N using the spring scale. Measure with the meter-stick the corresponding elongations x1, x2,...x10. Make a graph having on the x axis the elongations (x1, x2,...,x10) and on the y axis the forces (1, 2, ..10 N). Draw a fit line (least squares) through all 10 points and measure its slope k.

Then take all the rubber bands you have (let say 5 bands) and couple them in parallel. Fix one end of the bunch and stretch the other end with increasing forces of 1 N, 2 N,... ,10 N. Measure with the meter-stick the corresponding elongations x1', X2',..,X10'. Make the same graph as above, draw the same fit line between points and measure again its slope k'.

If James is right you should obtain for the k' the EXACT value

`k' = k/5`

If Larry is right you should obtain a value for k' DIFFERENT from the above value (but slightly close to k/5).

Observation:

The experiment can be repeated by coupling 2 by 2 (in pairs) the bands and doing the same measurements as above.

If Larry is right you should obtain FOR ALL PAIRS the EXACT value for each pair k'.

`k' =k/2`

This experiment takes a bit longer but the results are more accurate.

I need to make a small correction. The results from parallel coupling need to be exchanged with the results to series coupling. Therefore the EXACT value for k' when 5 bands (respectively 2 bands) are in parallel is

(respectively ).

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