Respuesta :
Answer:
Option 4. 0.05 J
Solution:
As per the question:
Spring constant, k = 10 N/m
Equilibrium position, x = 0.1 m
Now,
The potential energy of the spring is given by:
[tex]U = \frac{1}{2}kx^{2}[/tex]
And also from the principle of conservation of energy:
KE = U (1)
where
KE = Maximum Kinetic Energy
U = Potential energy
Thus
KE = U = [tex]\frac{1}{2}kx^{2}[/tex]
KE = U = [tex]0.5\times 10\times 0.1^{2} = 0.05\ J[/tex]
Answer:4
Explanation:
Given
spring constant [tex]k=10 N/m[/tex]
if it is stretched 0.1 m from equilibrium position
If the spring is at maximum extension then the Elastic Potential Potential energy is maximum and is equal to Total energy of the system
When the spring is at equilibrium , total energy will be equal to kinetic energy
So maximum kinetic Energy is equal to
[tex]=\frac{1}{2}kx^2[/tex]
[tex]=\frac{1}{2}\times 10\times (0.1)^2[/tex]
[tex]=0.05\ J[/tex]
