Question

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?

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.

I don't need the answer the correct answer is D

Answer 1

Answer:

D on ED

Explanation:

Answer 2

We can calculate the length of each spring by using the relationship:
$F=kx$
where
F is the force applied to the spring
k is the spring constant
x is the length of the spring (measured with respect to its rest position)

Re-arranging the equation, we have
$x= \frac{F}{k}$

The force applied to both spring is F=60 N. Spring A has spring constant of k=4 N/m, therefore its length with respect to its rest position is
$x_A= \frac{F}{k_A}= \frac{60 N}{4 N/m}=15 m$
Spring B has spring constant of k=5 N/m, so its length with respect to its rest position is
$x_B= \frac{F}{k}= \frac{60 N}{5 N/m}=12 m$

Therefore, the correct answer is
D.Spring A is 3 m longer than spring B because 15 – 12 = 3.