Question

25points
derive

a spring of spring constant K is attached to A block resting on a smooth horizontal surface. The Other end the spring is attached to a rigid wall the spring is light and massless. it is stretched to a displacement X. "find an expression for potential energy of the spring".​

Answer 1

Answer:

The spring constant is the measure of stiffness of a spring. Hooke's law gives us the force we need to find elastic potential energy. Looking at a graph of force versus displacement, we can find that the formula for elastic potential energy is PE = 1/2(kx^2).

Answer 2

according to hookes law

force required to compress or expand a spring is...-kx and kx respectively.....(k: spring constant...x: length of expansion of spring..)

so fs=kx

we know that work is the, product of force and distance

see the attachment...now...

now let's do it by second, method.....XD

since,by hookes law

we know fs=kx

this is in the form of y=mx,,which means that,it will pass through the origin....

now let's draw the graph of,force and distance....

we know that area under the slope of distance- force graph,is the work done

so, seeing from the graph (see the attachment)

work=area of triangle formed=1/2 kx.x=1/2kx^2

and we know that total work done is the total energy,consumed or absorbed

so PE=1/2kx^2