Respuesta :
Answer:
Elastic potential energy, E = 200 J
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 :
[tex]E=\dfrac{1}{2}k(x+x')^2[/tex]
[tex]E=\dfrac{1}{2}\times 4\times (10)^2[/tex]
E = 200 J
So, the elastic energy in the spring after Varg stretches the spring is 200 J. hence, this is the required solution.
Answer:
200 J
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
[tex]U = \frac{1}{2}kx^{2}[/tex]
U = 0.5 x 4 x 10 x 10
U = 200 J
Thus, the elastic potential energy is 200 J.
