Question

37. One spring has force constant 200 Nm-*
another has force constant 500 Nm-*. If they
are joined in series, the force constant will be
nearest to
A. 700 Nm
B. 300 Nm
e. 143 Nm
D. 100 Nm
with solution and plz.​

Answer 1

Answer:

143 Nm as solved in image below

Explanation:

Answer 2

Answer:

One spring has a force constant of  200 Nm another has a force constant of 500 Nm. If they are joined in series, the force constant will be $143Nm$

The correct answer is option e

Explanation:

Let $k_1$ be the force constant of the first spring and $k_2$ be the force constant of the second spring.

Since they are connected in series, the equivalent spring constant will be given by,  

$\frac{1}{k_{equ} } = \frac{1}{k_1} +\frac{1}{k_2}$      ...(1)

In the question, it is given that,

$k_1 = 200Nm$          $k_2 = 500Nm$

Substituting the above values into equation (1),

$\frac{1}{k_{equ} } = \frac{1}{200} +\frac{1}{500}$

       $= \frac{7}{1000}$  

$k _{equ} = \frac{1000}{7}$

        $=142.8 Nm$  ≈ $143Nm$

The correct answer is option e