Two massless springs (S1 and S2) are arranged such that one hangs vertically downward and the other is vertically upward, as shown in figure (a). When a 0.275-kg mass is suspended from S1, it stretches by an amount Δx1 = 0.062 m, as shown in figure (b). Spring S1 is now lowered so that the mass rests on and compresses spring S2, as shown in figure (c). If S2 has a spring constant

Question

k2 = 94.0 N/m,determine the amount S1 is stretched (Δx1s) when the elastic potential energy of the two springs is the same. Use
g = 9.80 m/s2 for the magnitude of the acceleration due to gravity.

Answer 1

From the given data, we can find the spring constant (k1) of S1.

The force (F) exerted by the spring when the mass is suspended is given by:

F = k1 * Δx1

The force acting on the mass is also equal to the weight of the mass:

F = m * g

So we can write the equation as:

k1 * Δx1 = m * g

Plugging in the given values, we get:

k1 * 0.062 m = 0.275 kg * 9.80 m/s²

Solving for k1, we get:

k1 = (0.275 * 9.80) / 0.062 = 43.4 N/m

Now, we need to find the amount S1 is stretched (Δx1s) when the elastic potential energy of the two springs is the same.

The elastic potential energy (PE) of a spring is given by the formula:

PE = (1/2) * k * (Δx)²

We need to find the total elastic potential energy of the system when the elastic potential energy of the two springs is equal. Therefore, we have:

(1/2) * k1 * (Δx1s)² = (1/2) * k2 * (Δx1 - Δx1s)²

Solving for Δx1s, we get:

(43.4 N/m) * (Δx1s)² = (94.0 N/m) * [0.062 m - Δx1s)]²

We have a quadratic equation in terms of Δx1s. Dividing both sides by (1/2), we get:

(43.4) * (Δx1s)² = (94.0) * [0.062 - Δx1s]²

Expanding the expression, we get:

43.4 * Δx1s² = 94.0 * (0.003844 - 0.124Δx1s + Δx1s²)

Now, we simplify the expression:

43.4 * Δx1s² = 94.0 * 0.003844 - 94*0.124Δx1s + 94.0Δx1s²

-50.6Δx1s² + 11.696Δx1s - 0.36176 = 0

To solve for Δx1s, we can use the quadratic formula:

Δx1s = [-B + sqrt(B² - 4AC)] / 2A or [-B - sqrt(B² - 4AC)] / 2A

Where A = -50.6, B = 11.696, and C = -0.36176

Plugging in the values, we get:

Δx1s = [-11.696 + sqrt(11.696² - 4*(-50.6)*(-0.36176))] / 2 * -50.6

or

Δx1s = [-11.696 - sqrt(11.696² - 4*(-50.6)*(-0.36176))] / 2 * -50.6

Calculating the two possible values for Δx1s:

Δx1s ≈ 0.0384 m or Δx1s ≈ 0.245 m

However, since the mass is only suspended a distance of 0.062 m for spring S1, the value of 0.245 m is not possible. Therefore, the amount S1 is stretched when the elastic potential energy of the two springs is the same is:

Δx1s ≈ 0.0384 m