it takes 80 j of work to stretch a spring 0.5 m from its equilibrium position. how much work is needed to stretch it an additional 0.5m

a)240 j b)300 j c) 400j d)500j

Question

it takes 80 j of work to stretch a spring 0.5 m from its equilibrium position. how much work is needed to stretch it an additional 0.5m

a)240 j   b)300 j   c) 400j   d)500j

anonymous · Answer

HI :D

ganeshie8 · Answer

$\large   W =    -\frac{1}{2} k x^2 $

plugin x = 0.5 and find $k$ first

ganeshie8 · Answer

$\large   80 =    -\frac{1}{2} k 0.5^2$

solve $k$

anonymous · Answer

=12.64 or =4square root of 10

ganeshie8 · Answer

nope, try again :)

ganeshie8 · Answer

$80 =    -\frac{1}{2} k 0.5^2 $

multiply 2 both sides


$160 =  -k 0.5^2$

divide 0.5^2 both sides

$160/0.5^2   = -k$

$k = -640$

ganeshie8 · Answer

So,

$W = \frac{1}{2}640\times x^2  =   320 \times  x^2$

plugin x = 1 to find the total work done in streteching from equilibrium to 1 meter

$W =    320 \times   1^2  = 320$

since we wanto knw the work done in moving from 0.5 to 1m
subtract 80J from this :-
$320 - 80 = 240J$

anonymous · Answer

thanks so much this was a lot of help!

ganeshie8 · Answer

np :)