From the slope of the graph, find the mass of the system and compare it to the expected value of 4kg using % Difference. This involves a derivation – show how you get the mass from the slope. Experimental Acceleration vs Accelerating Force8y = 1.0107x.esi711234567Accelerating Forceco6.st2.Experimental Acceleration Eq 1.Eq 2.EFEq 3.ΣεF = Whanging= m19m19at?1аехресtedAy =MM(1kg)(9.8)/1kg+3kg =2.45 m/s^2(2kg)(9.8)/2kg+2kg =4.90 m/s^21kg(9.8) = 9.8 N2(2.02m)/(1.28s)^2= 2.47 m/s^22(2.02m)/(0.91s)^2= 4.89 m/s^2Run 1Run 22kg(9.8) = 19.6 N%3!Run 33kg(9.8) = 29.4 N(3kg)(9.8)/3kg+1kg =7.35 m/s^22(2.02m)/(0.74s)^2= 7.38 m/s^2%3D%3!

Answered: Experimental Acceleration vs…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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From the slope of the graph, find the mass of the system and compare it to the expected value of 4kg using % Difference. This involves a derivation – show how you get the mass from the slope.

Experimental Acceleration vs Accelerating Force
8
y = 1.0107x.esi
7
1
1
2
3
4
5
6
7
Accelerating Force
co
6.
st
2.
Experimental Acceleration

Eq 1.
Eq 2.
EF
Eq 3.
Σε
F = Whanging
= m19
m19
at?
1
аехресted
Ay =
M
M
(1kg)(9.8)/1kg+3kg =
2.45 m/s^2
(2kg)(9.8)/2kg+2kg =
4.90 m/s^2
1kg(9.8) = 9.8 N
2(2.02m)/(1.28s)^2
= 2.47 m/s^2
2(2.02m)/(0.91s)^2
= 4.89 m/s^2
Run 1
Run 2
2kg(9.8) = 19.6 N
%3!
Run 3
3kg(9.8) = 29.4 N
(3kg)(9.8)/3kg+1kg =
7.35 m/s^2
2(2.02m)/(0.74s)^2
= 7.38 m/s^2
%3D
%3!

Expert Solution

Step 1

We are given the graph of acceleration vs accelerating force and the graph is a straight line as follows :

$y = 1.0107 x^{0.9951}$

Determine:

the mass of the system by using slope of the graph

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