A spring (spring 1) with a spring constant of 520N/m is attached to a wall and connected to another weaker spring (spring 2) with a spring constant of 270N/m on a horizontal surface. Then an external force of 70N is applied to the end of the weaker spring (#2).

how much potential energy is stored in each spring

F = k1*x1

70 = (520)*x1

F = k2*x2
70 = 270*x2

Plug x's into potential energy for a spring equation U = 0.5kx^2 for respective springs

Spring #1:

70N * 1m/520N. = 0.1346 m.
PE=0.5F * d = 0.5*70 * 0.1346 = 4.71 J.

Spring #2:
70N/270N * 1m. = 0.2593 m.
PE=0.5F * d = 0.5*70 * 0.2593 = 9.07 J.

To calculate the potential energy stored in each spring, we can use the formula:

Potential energy (U) = (1/2) * k * x^2

Where:
U: Potential energy
k: Spring constant
x: Displacement from the equilibrium position

For spring 1:
Spring constant (k1) = 520 N/m

Spring 1 is attached to a wall, which means it will not be displaced when the external force is applied to spring 2. Therefore, the potential energy stored in spring 1 is zero.

U1 = 0

For spring 2:
Spring constant (k2) = 270 N/m
External force (F) = 70 N

To find the displacement (x2) of spring 2, we can use Hooke's Law:

F = k2 * x2

Rearranging the equation:

x2 = F / k2
x2 = 70 N / 270 N/m
x2 = 0.259 m (rounded to three decimal places)

Now we can calculate the potential energy stored in spring 2:

U2 = (1/2) * k2 * x2^2
U2 = (1/2) * 270 N/m * (0.259 m)^2
U2 = 10.97 J (rounded to two decimal places)

Therefore, the potential energy stored in spring 1 is 0 J, and the potential energy stored in spring 2 is 10.97 J.

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