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Lab+5+Conservation+of+Energy (1) Microsoft Word Acrobat Title Subtitle Subtle En 1 Normal 1 No Spacing Heading 1...

Question

Lab+5+Conservation+of+Energy (1) Microsoft Word Acrobat Title Subtitle Subtle En 1 Normal 1 No Spacing Heading 1...

Lab+5+Conservation+of+Energy (1) Microsoft Word Acrobat Title Subtitle Subtle En 1 Normal 1 No Spacing Heading 1 Heading 2 Styles eriment 1: Work Done by a Spring Table 1: Spring Scale Force Data Distance, Forc@average ΔDistance Force (N)x (m) Ax (m)Work (u) 0 Click here to enter text 0.01 0.007) OUN 0.01 Click here to 0.01 enter text. 0.03 J 1.5N 0.02 Click here to enter text 0.066J ON 0.03 Clic here to0.010.124) 3.1N 0.04 click here to 0.01 0.2J 0.05 Note, you will finish completing Table 1 in the Post-Lab Questions section. -Lab Questions Create a Force vs. Displacement (stretch) graph. Insert photo of graph with your name clearly visible in the background:
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Lab+5+Conservation+of+Energy (1) Microsoft Word Acrobat Title Subtitle Subtle En 1 Normal 1 No Spacing Heading 1 Heading 2 Styles eriment 1: Work Done by a Spring Table 1: Spring Scale Force Data Distance, [email protected] ΔDistance Force (N)x (m) Ax (m)Work (u) 0 Click here to enter text 0.01 0.007) OUN 0.01 Click here to 0.01 enter text. 0.03 J 1.5N 0.02 Click here to enter text 0.066J ON 0.03 Clic here to0.010.124) 3.1N 0.04 click here to 0.01 0.2J 0.05 Note, you will finish completing Table 1 in the Post-Lab Questions section. -Lab Questions Create a Force vs. Displacement (stretch) graph. Insert photo of graph with your name clearly visible in the background:

Answers

Let us consider a spring having a spring constant k N/m

2)Here the Work done by the spring can be determined from the graph by finding the area under the graph

That is Work done =1/2*b*h= 1/2*0.04*3.3=0.066 J

3)The average force can be calculated using the equation F=-kx

where k is the spring constant, x is the strecth = 1cm =0.01m

Work done = 1/2 k x2 = 0.066 J

Therefore k= (0.066/0.0001)*0.5= 330 N/m

Force F= kx= 330* 0.01= 3.3N

4)Work done in each segment is given by W= 1/2 kx2

W1= 1/2 k x2 = 1/2*330*0.01 =0.0165J

W2=1/2 * 330*0.022=0.066J

W3=1/2 * 330*0.032=0.1485J

W4=1/2 * 330*0.042=0.264J

W5=1/2 * 330*0.052=0.4125J

For comparing these values with the obtained values in the post lab question2 ,take the difference between the two 5),Work done by a spring= \huge \int F.dx

F= 5.4- 2.3x2 +4.3 x3.

W=\huge \int5.4- 2.3x2 +4.3 x3 dx ; x=0.05m

5.4x-2.3x3/3+4.3 x4/4 =0.26904 J

1) Graph to calculate the workdone by a spring.

1 unit χ aus s o.oim. FoRcE l unit alona yaxs 3 2 an Are undea the bagh o.o66 Nm


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