Question

A spring of spring constant 5×10^3 is stretched initially by 5cm from the unstretched position.then the work required to stretch it furthur by another 5cm is

Answer 1

Here, W=1/2k(x22-x12)                = 1/2×5×103((10/100)2-(5/100)2)                =1/2×5×103/104(100-25)                =18.75J Hence the work required to stretch the spring further by another 5cm is 18.75 N-m Hope it helps  Thanks

Answer 2

Answer:

The work required to stretch the spring further by another 5 cm is 18.75 J.

Explanation:

Given that,

Spring constant $k = 5\times10^{3}$

Initially stretched = 5 cm

Again stretched = 5 cm

The work done is defined as

$W= \dfrac{1}{2}k(x_{2}^2-x_{1}^2)$

Here, k = spring constant

x= stretched

$W=\dfrac{1}{2}\times5\times10^{3}((\dfrac{10}{100})^2-(\dfrac{5}{100})^2)$

$W= 18.75\ J$

Hence, The work required to stretch the spring further by another 5 cm is 18.75 J.