An elastic spring of unstretched length L and | vedclass

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

Question

(D) The work done in stretching a spring by an extension $x$ is given by $W = \frac{1}{2} k x^2$. Initially, the spring is stretched by $x$. The poten

Vedclass · Accepted Answer

$\frac{k y}{2}(2 x+y)$

Vedclass · Answer

(D) The work done in stretching a spring by an extension $x$ is given by $W = \frac{1}{2} k x^2$. Initially, the spring is stretched by $x$. The potential energy stored is $U_1 = \frac{1}{2} k x^2$. Then, it is further stretched by $y$, so the total extension becomes $(x + y)$. The potential energy stored is $U_2 = \frac{1}{2} k (x + y)^2$. The work done during the second stretching is the change in potential energy: $W = U_2 - U_1 = \frac{1}{2} k (x + y)^2 - \frac{1}{2} k x^2$. Expanding the term $(x + y)^2$: $W = \frac{1}{2} k (x^2 + y^2 + 2xy - x^2) = \frac{1}{2} k (y^2 + 2xy)$. Factoring out $y$: $W = \frac{ky}{2} (2x + y)$.

An elastic spring of unstretched length $L$ and force constant $k$ is stretched by a small length $x$. It is further stretched by another small length $y$. Work done during the second stretching is

Similar Questions

When a spring is stretched by $2 \, cm$,it stores $100 \, J$ of energy. If it is stretched further by $2 \, cm$,the stored energy will be increased by ............. $J$.

$A$ block of mass $m$ moving at a speed $v$ compresses a spring through a distance $x$ before its speed becomes one fourth. Find the spring constant of the spring.

$A$ spring with spring constant $k$ is extended from $x = 0$ to $x = x_1$. The work done will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module