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.

(a) how much work is needed to stretch the spring from 32 cm to 34 cm? (round your answer to two decimal places.)

(b) how far beyond its natural length will a force of 25 n keep the spring stretched? (round your answer one decimal place.)

Answer

given,

work = 5 J

spring stretch form 30 cm to 39 cm

x = 0.39 - 0.30 = 0.09 m

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 = 0.741 J

b) F = k x

  25 = 1234.568 × x

     x = 0.0205 m

     x = 2.05 cm

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl