Question
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
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.
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.
Answers
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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
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