suppose a .3 kg mass on a spring that has been compress .1 m has elastic potential energy of 1 j. how much further must the spring be compressed to triple the elastic potential energy?

the answer is .07, but how do i get that?

elastic PE=1/2 k x^2 where x is the total distance. figure k from the first sentence..I assume the .3kg is compressing the spring due to its weight.

1j=mg*distance=1/2 kx^2

k= 2j/.1^1= 200j/m^2

triple energy?
3j=1/2 (200j/m^2)x^2 solve for x.

Suppose a 0.30-kg mass on a spring that has been compressed 0.10 m has elastic potential energy of 1.5 J. What is the spring constant?

Select one:
a. 300 N/m
b. 20 N/m
c. 200 N/m
d. 10 N/m

To find out how much further the spring must be compressed to triple the elastic potential energy, we can use the formula for elastic potential energy:

Elastic Potential Energy = 0.5 * k * x^2

Where:
- Elastic Potential Energy (E) is given as 1 J.
- The mass (m) is 0.3 kg.
- The spring constant (k) is unknown.
- The original compression of the spring (x) is 0.1 m.

We want to triple the elastic potential energy, so let's call the new compression distance y.

Triple the Elastic Potential Energy = 3 * Elastic Potential Energy

Substituting the given values:

3 * 1 J = 0.5 * k * y^2

Now, we can solve for the spring constant (k) using the given mass (m):

k = (4 * pi^2 * m) / T^2

Where:
- pi is approximately 3.14159
- T is the period of oscillation of the mass, which depends on the spring constant and mass.

Since we don't know the period (T), let's consider the fact that the period does not change when we triple the elastic potential energy. Therefore, the ratio of the spring compression squared must also be tripled:

(y^2) / (x^2) = 3

Substituting the given values:

(y^2) / (0.1^2) = 3

Simplifying the equation:

y^2 = 0.03

Taking the square root of both sides:

y = sqrt(0.03)

y ≈ 0.173205

Therefore, the spring must be compressed by approximately 0.173205 m or 0.17 m (rounded to two decimal places) further to triple the elastic potential energy.

two pellets each with a charge of 1 microcoulomb 10 to the negative 6 c are located 3 cm apart what is the electrical force between them what would be the mass of a object that would experience the same force in the earths gradational field