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a spring of force constant k is first stretched by distance a from its natural length and then further by distance b the work done in stretching the part b is?? answer is 1/2 Kb(2a+b)

Vaibhav , 9 Years ago
Grade 11
anser 3 Answers
DR STRANGE
Hi Vaibhav,
the work done when a spring is stretched by distance x from its natural length is,
 \frac{1}{2}\times K\times x^{2}
for distance a its,
 \frac{1}{2}\times K\times a^{2}
for as it is again stretched by distance b,
total work done is,
\frac{1}{2}\times K\times (a+b)^{2}
for the work done in stretching the part b is just,
 \frac{1}{2}\times K\times (a+b)^{2}-\frac{1}{2}\times K\times (a)^{2} 
=\frac{1}{2}\times K\times b (2a+b)^{2}
I hope you will get it,
Plz approve if you are satisfied :)
ApprovedApproved
Last Activity: 9 Years ago
DR STRANGE
sorry print mistake ans is
=\frac{1}{2}\times K\times b (2a+b)

Last Activity: 9 Years ago
Priyanshi srivastava
Ws= -1/2 k(x2f - x2i)
∆U= -Ws
∆U = 1/2 k (x2f - x2i)
U2 - U1 =  1/2 kx22 - 1/2 kx21
If at x1=0   U1 =0 &.                                                                                
    at  x2=x   U2=U . 
Then U=1/2 kx2
Last Activity: 7 Years ago
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