Question

Varg sees a spring that has a spring constant of 4 N/m that is stretched 5 m. He stretches the spring an additional 5 m. Consider the system to be the spring. How much elastic energy does the spring have after Varg stretches the spring?

Answer 1

200 J

Step-by-step explanation:

Spring constant, K = 4 N/m

initial extension, x' = 5 m

Another extension, x'' = 5 m

Total extension, x = 5 + 5 = 10 m

The elastic potential energy is given by

`U = (1)/(2)kx^(2)`

U = 0.5 x 4 x 10 x 10

U = 200 J

Thus, the elastic potential energy is 200 J.

Answer 2

Elastic potential energy, E = 200 J

Step-by-step explanation:

It is given that,

Spring constant, K = 4 N/m

initial stretching in the spring, x = 5 m

Finally, it is stretched an additional 5 m i.e. x' = 5 m

Let E is the elastic energy in the spring after Varg stretches the spring. it is given by :

`E=(1)/(2)k(x+x')^2`

`E=(1)/(2)* 4* (10)^2`

E = 200 J

So, the elastic energy in the spring after Varg stretches the spring is 200 J. hence, this is the required solution.