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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)


4 years ago

DR STRANGE
212 Points
							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 :)

4 years ago
DR STRANGE
212 Points
							sorry print mistake ans is
$=\frac{1}{2}\times K\times b (2a+b)$


4 years ago
Priyanshi srivastava
15 Points
							Ws= -1/2 k(x2f - x2i)∆U= -Ws∆U = 1/2 k (x2f - x2i)U2 - U1 =  1/2 kx22 - 1/2 kx21If at x1=0   U1 =0 &.                                                                                    at  x2=x   U2=U . Then U=1/2 kx2

2 years ago
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