A spring of original length L and spring cons | vedclass

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

Question

(D) The potential energy stored in a spring stretched by a length $x$ is given by $U_1 = \frac{1}{2}Kx^2$.When the spring is further stretched by an a

Vedclass · Accepted Answer

$\frac{1}{2}Ky(2x + y)$

Vedclass · Answer

(D) The potential energy stored in a spring stretched by a length $x$ is given by $U_1 = \frac{1}{2}Kx^2$.When the spring is further stretched by an additional length $y$,the total extension becomes $(x + y)$.The total potential energy stored in the spring is $U_2 = \frac{1}{2}K(x + y)^2$.The work done in the second stretching is the change in potential energy:$W = U_2 - U_1$$W = \frac{1}{2}K(x + y)^2 - \frac{1}{2}Kx^2$$W = \frac{1}{2}K(x^2 + 2xy + y^2) - \frac{1}{2}Kx^2$$W = \frac{1}{2}K(x^2 + 2xy + y^2 - x^2)$$W = \frac{1}{2}K(2xy + y^2)$$W = \frac{1}{2}Ky(2x + y)$.

$A$ spring of original length $L$ and spring constant $K$ is stretched by a small length $x$. It is further stretched by another small length $y$. Find the work done in the second stretching.

Similar Questions

$A$ mass of $0.5\,kg$ moving with a speed of $1.5\,m/s$ on a horizontal smooth surface,collides with a nearly weightless spring of force constant $k = 50\,N/m$. The maximum compression of the spring would be ............. $m$.

When a body of mass $m$ is attached to a vertical spring of natural length $L$ and released,the spring stretches by a distance $h$. What will be the potential energy of the stretched spring?

Does the energy stored in a spring change when it is stretched or compressed?

Is the work done by the stretching force in a stretched spring positive or negative?

Vedclass Test Series

Exam Paper Generator

Online Exam Module