Question

Obtain an expression for the potential energy!
of a spring having spring constant k when it
is extended through a length x. ​

Answer 1

Answer:

$hllw$

Explanation:

Since workdone on the object is equal to mgh, an energy equal to mgh units is gained by the object . This is the potential energy (Ep ) of the object. It is clear that the work done is equal to the change in the kinetic energy of an object. If u = 0, the work done will be W =12mv2 W = 1 2 m v 2.....

Hope it's help u

Answer 2

Answer:

The potential energy of a spring having spring constant k when it is extended through a length x. is given by ​$P.E = \frac{1}{2}kx^2$

Explanation:

Hooke's laws

The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring for short distances. Since this restoring force is in the opposite direction, a negative sign is used.

If x is the displacement relative to the unstretched length of the spring and F is the force exerted by it. Then,

$F_{s}=-k x$

Here,

$F_{s}$ is the force of the spring

$x$is the unstretched length of the spring

$k$ is the spring constant?

The area under the curve = Area of the shaded region of the curve

                                        $= \frac{1}{2} \times base \times height\\= \frac{1}{2} \times x \times (kx)\\= \frac{1}{2}kx^2q$

Both of the approaches give the same answer.

Thus, elastic potential energy stored in spring with “x” displacement is given by,

$P.E = \frac{1}{2}kx^2$