Question

Suppose that 5 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 39 cm.

Answer 1

Answer

given,

work = 5 J

spring stretch form 30 cm to 39 cm

$W = \dfrac{1}{2}kx^2$

x = 0.39 - 0.30 = 0.09 m

$5 = \dfrac{1}{2}k\times 0.09^2$

$k = \dfrac{5\times 2}{0.09^2}$

k = 1234.568 N/m

a) work when spring is stretched from  32 cm to 34 cm

x₂= 0.34 -0.30 = 0.04 m

x₁ = 0.32 - 0.30 = 0.02

$W = \dfrac{1}{2}k(x_2^2-x_1^2)^2$

$W = \dfrac{1}{2}\times 1234.568 \times (0.04^2-0.02^2)^2$

W = 0.741 J

b) F = k x

  25 = 1234.568 × x

     x = 0.0205 m

     x = 2.05 cm