It takes 5.06 j of work to stretch a hooke’s-law spring 13.3 cm from its unstressed length. how much the extra work is required to stretch it an additional 13.4 cm? answer in units of j. it takes 5.06 j of work to stretch a hooke’s-law spring 13.3 cm from its unstressed length. how much the extra work is required to stretch it an additional 13.4 cm? answer in units of j.

Answers:

extra work required = 0.038 J

Explanation:

let X = 13.3 cm and Y = 13.4 cm, W1 be the work done to stretch the spring 13.3 cm and W2 be the work done to stretch the spring 13.4cm .let F be the force stretching the spring.

from Work and Energy principles:

W1 = F×X×cos(Ф) = F×X , since the force is acting in the direction at which the             spring is stretching , cos(Ф) = 1

then :

W1 = F×X

F = W1/X

  = 5.06/(13.3×10^-2)

  = 38.05 J

and,

W2 = F×Y×cos(Ф) = F×Y ,  ince the force is acting in the direction at which the             spring is stretching , cos(Ф) = 1

W2 = F×Y

      =  (38.5)×(13.4×10^*2)

      = 5.098 J

therefore, the extra required is 5.098 - 5.06 = 0.038 J.

its a 5.5 n

glad that

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west