Question

The force constant of a spring is 400 N/m. How much work must be done on the spring to stretch it (a) by 6.0 cm (b) from x = 4.0 cm to x = 6.0 cm, where x = 0 is the relaxed position of the spring

Answer 1

work done = force×displacement
so now in case of spring -kx is the workdon in which k is force constant and x is the displacement now
a. work done= - 400× 6 = -2400 joule
b.here in x =4cm it would already a wrkdone of 400×4 =1600 joule plus again stretched work of 6-4= 2 cm so net wrkdone of 2×400 =800 so 800+1600=2400joule

Answer 2

work done by spring force = 1/2Kx^2
where K is spring constant

1) work done =1/2 x 400 x (6/100)^2 Joule = 1/2 x 400 x 36 x 10^-4 Joule
=72 x 10^-2 Joule
=0.72 Joule

2) U = -K. xdx
integrate within limit x =6 to x =4
U = -K /2[x^2] limit [4, 6]
=-200[36 -16] x 10^-4 Joule
= -200 x 20 x 10^-4 Joule
= -0.4 Joule
we know ,
negative of potential energy is equal to work done
e.g W = -U
= -(-0.4) =0.4 Joule