Question

1. Suppose the spring in Sample Problem A is replaced with a spring that stretches

36 cm from its equilibrium position.

a. What is the spring constant in this case?

b. Is this spring stiffer or less stiff than the one in Sample Problem A?

Answer 1

Answer:

a. spring constant = 125 N/m

b. This spring is less stiff than the one in Sample Problem A.

Explanation:

P.S - The Sample Problem A is as follows :

Given - Sample Problem A - A load of 45 N attached to a spring that is

            hanging vertically stretches the spring 0.14 m. What is the spring

             constant?

            Suppose the spring in Sample Problem A is replaced with a spring  

            that stretches 36 cm from its equilibrium position.

To find - a. What is the spring constant in this case?

            b. Is this spring stiffer or less stiff than the one in Sample Problem A.

Proof -

As given,

Load = 45 N

Amplitude = 0.14 m

Let the spring constant = k

As we know that,

Load = k (Amplitude)

⇒45 = k(0.14)

⇒k = $\frac{45}{0.14}$ = 321.43

∴ we get

Spring constant in Sample problem A = 321.43

Now,

a.)

Given, Amplitude = 36 cm = 0.36 m

Let the spring constant = k₁

⇒45 = k₁ (0.36)

⇒k₁ = $\frac{45}{0.36}$ = 125 N/m

b.)

AS we can see that k₁  < k

⇒ This spring is less stiff than the one in Sample Problem A.

Answer 2

a. spring constant = 125 N/m

b. This spring is less stiff than the one in Sample Problem A.

what is spring constant?

The spring constant generally shows the stiffness of the spring and is the ratio of the force applied to the deflection of the spring.

It is given in the question that:

Sample Problem A - A load of 45 N attached to a spring that is

hanging vertically stretches the spring 0.14 m.  

         

Suppose the spring in Sample Problem A is replaced with a spring  

that stretches 36 cm from its equilibrium position.

a. What is the spring constant in this case?

As given,

Load F = 45 N

Amplitude  x= 0.14 m

Let the spring constant = k

As we know that spring force will be

$F=k\times x$

$45=k\times 0.14$

⇒k = 321.43N/m

∴ we get

Spring constant in Sample problem A is   $k=321.43\ \frac{N}{m}$

Now,

Given, Amplitude = 36 cm = 0.36 m

Let the spring constant = k₁

⇒45 = k₁ (0.36)

⇒k₁= 125 N/m

b.) Is this spring stiffer or less stiff than the one in Sample Problem A.

AS we can see that k₁  < k  new spring is less stiff than the one in Sample Problem A.

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