When a block of mass M is suspended by a long wire of length L. the length of the wire becomes (L+ l). The elastic potential energy stored in the extended wire is:

Question

When a block of mass M is suspended by a long wire of length L. the length of the wire becomes (L+ l). The elastic potential energy stored in the extended wire is: (A) (1/2)Mgl (B) (1/2)MgL (C) mgl (D) MgL

Tardigrade Admin · Accepted Answer

U = (1/2) (work done by gravity) U = (1/2) Mgl

Q. When a block of mass $M$ is suspended by a long wire of length $L$ . the length of the wire becomes $(L + l)$ . The elastic potential energy stored in the extended wire is:

$\frac{1}{2} M g l$

$\frac{1}{2} M gL$

$m g l$

$M gL$

$U = \frac{1}{2}$ (work done by gravity)
$U = \frac{1}{2} M g l$

Q. When a block of mass M is suspended by a long wire of length L. the length of the wire becomes (L+l). The elastic potential energy stored in the extended wire is:

21​Mgl

21​MgL

mgl

MgL