A mechanical system consists of two springs w | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) When two springs with stiffness constants $k_1$ and $k_2$ are connected in series, the equivalent spring constant $K_{	ext{eq}}$ is given by the

Vedclass · Accepted Answer

$\frac{1}{2}\left(\frac{k_1 k_2}{k_1+k_2}\right) \Delta^2$

Vedclass · Answer

(A) When two springs with stiffness constants $k_1$ and $k_2$ are connected in series, the equivalent spring constant $K_{	ext{eq}}$ is given by the formula: $\frac{1}{K_{	ext{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{k_1 + k_2}{k_1 k_2}$ Therefore, $K_{	ext{eq}} = \frac{k_1 k_2}{k_1 + k_2}$. The work done $W$ to stretch a spring system by a total displacement $\Delta$ is equal to the potential energy stored in the equivalent spring: $W = \frac{1}{2} K_{	ext{eq}} \Delta^2$ Substituting the value of $K_{	ext{eq}}$: $W = \frac{1}{2} \left( \frac{k_1 k_2}{k_1 + k_2} ight) \Delta^2$.

$A$ mechanical system consists of two springs with stiffness coefficients $k_1$ and $k_2$ connected in series. The minimum work to be performed on the system to stretch it by a total displacement $\Delta$ is

Similar Questions

Find the maximum tension in the spring if the spring is initially at its natural length when the block of mass $m$ is released from rest.

Explain the elastic potential energy of a spring and obtain an expression for this energy.

$A$ sphere of mass $25 \,g$ is placed on a vertical spring. The spring is compressed by $0.2 \,m$ using a force of $5 \,N$. When the spring is released, the mass reaches a maximum height of (take $g = 10 \,m/s^2$):

$A$ spring of force constant $800\, N/m$ has an extension of $5\, cm$. The work done in extending it from $5\, cm$ to $15\, cm$ is ............. $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module