MATH SOLVE

2 months ago

Q:
# A force of 60 N is used to stretch two springs that are initially the same length. Spring A has a spring constant of 4 N/m, and spring B has a spring constant of 5 N/m.How do the lengths of the springs compare?A:Spring B is 1 m longer than spring A because 5 – 4 = 1.B:Spring A is the same length as spring B because 60 – 60 = 0.C:Spring B is 60 m longer than spring A because 300 – 240 = 60.D:Spring A is 3 m longer than spring B because 15 – 12 = 3.

Accepted Solution

A:

Answer:D:Spring A is 3 m longer than spring B because 15 – 12 = 3.Step-by-step explanation:In this question, you should remember the Hooke's Law in physics.The Hooke's Law simply explains that the extension that occurs on a spring is directly proportional to the load applied on it.The mathematical expression for this law is[tex]F=-kx[/tex]where;F= force applied on the springx = the extension on the spring k= the spring constant which varies in spring.The question will need you to calculate the extension on the springs A and B then compare the values obtained.In spring AForce, F=60N and spring constant ,k=4 N/mTo find the extension x apply the expression;[tex]F=-kx\\\\60=-4*x\\\\60=-4x\\\\\frac{60}{-4} =\frac{-4x}{-4} \\\\\\-15=x[/tex]Here the spring extension is 15 mIn spring BForce, F=60N and spring constant , k=5N/mTo find the extension x apply the same expression[tex]F=-kx\\\\60=-5*x\\\\60=-5x\\\\\\\frac{60}{-5} =\frac{-5x}{-5} \\\\\\-12=x[/tex]Here the extension on the spring is 12 mCompareThe extension on spring A is 3 m longer than that in spring B because when you subtract the value of spring B from that in spring A you get 3m[tex]=15m-12m=3m[/tex]