
#1
Jan1106, 10:20 PM

P: 176

A spring with a force constant of 50 N/m is stretched continuously from x_{0}=0 to x_{2}=20 cm. Is more work required to stretch the spring from x_{1}=10 cm to x_{2}=20 cm than from x_{0} to x_{1}? Justify your answer mathematically.
I thought the work required is the same because the displacements are equal for both work equations... but I think I'm wrong. 



#2
Jan1106, 10:25 PM

P: 227

work = 1/2 kx^2
so then the difference would be greater between 20cm and 10cm w= EPEfEPEi = 0.5(k)4000.5(k)100 = 150k w0>1 = 0.5(k)(100)  0.5k(0) = 50k 



#3
Jan1106, 10:27 PM

P: 176

but wouldn't it be:
work1 = 1/2 k (10cm)^2 work2 = 1/2 k (20cm  10cm)^2=work1 ? 



#4
Jan1106, 10:57 PM

P: 227

is more work required for this spring?
no, the work is the difference between energies.
[tex] EPE_i + W = EPE_f [/tex] [tex] W = EPE_f  EPE_i [/tex] [tex] W = 0.5kx_f^2  0.5kx_i^2 [/tex] 



#5
Jan1106, 11:08 PM

P: 176

I remember now...
what does EPE mean? 



#6
Jan1106, 11:18 PM

P: 24

elastic potential energy



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