Spring Forces
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AP Physics C: Mechanics › Spring Forces
A $0.250\ \text{kg}$ block is attached to two ideal springs on a horizontal frictionless table. Spring 1 has constant $k_1 = 120\ \text{N/m}$ and Spring 2 has constant $k_2 = 80.0\ \text{N/m}$. The springs are connected in parallel between a wall and the block so that both springs stretch or compress by the same amount x when the block is displaced. The block is pulled a small distance and released, oscillating about equilibrium.
Given values:
- $k_1 = 120\ \text{N/m}$
- $k_2 = 80.0\ \text{N/m}$
- $m = 0.250\ \text{kg}$
Forces and model: For a displacement x, each spring exerts a restoring force $F_{s1} = -k_1 x$ and $F_{s2} = -k_2 x$. The net restoring force is $$F_{\text{net}} = -(k_1 + k_2)x,$$ so Newton’s Second Law becomes $$m,\frac{d^2x}{dt^2} = -(k_1+k_2)x.$$
Refer to the system described above. What is the effective spring constant for the system?
$200\ \text{N/m}$
$150\ \text{N/m}$
$48\ \text{N/m}$
$96\ \text{N/m}$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding spring forces in parallel configurations through the principle of force addition. When springs are connected in parallel, they experience the same displacement but exert independent forces that add, resulting in an effective spring constant keff = k₁ + k₂. In this problem, two springs with constants k₁ = 120 N/m and k₂ = 80.0 N/m are connected in parallel to a 0.250 kg block on a frictionless surface. Choice B is correct because for parallel springs, keff = k₁ + k₂ = 120 + 80.0 = 200 N/m, as both springs contribute their full restoring force at any displacement. Choice C (96 N/m) might result from incorrectly averaging the spring constants. To help students: Emphasize that parallel springs share displacement but add forces. Practice distinguishing between parallel (forces add) and series (displacements add) configurations.
A $2.0\ \text{kg}$ cart is attached to two springs connected in series along a horizontal frictionless track. Spring A has constant $k_A = 300\ \text{N/m}$ and spring B has constant $k_B = 600\ \text{N/m}$. The left end of spring A is fixed to a wall, spring A connects to spring B, and spring B connects to the cart. When the cart is displaced to the right from equilibrium by a small distance, both springs stretch, and each spring exerts a restoring force. For springs in series, the same force magnitude acts through both springs, and the total extension is the sum of individual extensions. The effective spring constant satisfies $\tfrac{1}{k_{\text{eff}}} = \tfrac{1}{k_A} + \tfrac{1}{k_B}$. The cart oscillates with small amplitude so that Hooke’s law applies.
Given values:
- Mass: m = $2.0\ \text{kg}$
- Spring constants: $k_A = 300\ \text{N/m}$, $k_B = 600\ \text{N/m}$
- Track is horizontal and frictionless
Refer to the system described above. What is the effective spring constant for the system?
$100\ \text{N/m}$
$900\ \text{N/m}$
$450\ \text{N/m}$
$200\ \text{N/m}$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding spring forces in series configuration and calculating effective spring constants. When springs are connected in series, they experience the same force but have different displacements, and the total displacement is the sum of individual displacements, leading to the reciprocal formula for effective spring constant. For springs in series: 1/keff = 1/kA + 1/kB = 1/300 + 1/600 = 2/600 + 1/600 = 3/600 = 1/200. Choice B is correct because keff = 200 N/m. Choice C (450 N/m) is incorrect as it might result from averaging the spring constants arithmetically instead of using the proper series formula. To help students: Derive the series formula from F = kx applied to each spring with the same F but different x values, and contrast with parallel springs. Watch for: using the parallel formula (simple addition) for series springs or arithmetic/algebraic errors in fraction manipulation.
A $0.600\ \text{kg}$ block is connected to two ideal springs in series on a frictionless horizontal surface. Spring 1 has constant $k_1 = 300\ \text{N/m}$ and Spring 2 has constant $k_2 = 150\ \text{N/m}$. The left end of Spring 1 is attached to a wall, Spring 1 connects to Spring 2, and the right end of Spring 2 attaches to the block. The block is displaced slightly and released, oscillating with small amplitude.
Given values:
- $k_1 = 300\ \text{N/m}$
- $k_2 = 150\ \text{N/m}$
- $m = 0.600\ \text{kg}$
Forces and model: In series, both springs carry the same force magnitude F while their extensions add: $x = x_1 + x_2$. Hooke’s Law gives $F = k_1 x_1 = k_2 x_2$, so the equivalent spring constant satisfies $$\frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2}.$$ The motion then satisfies $m,d^2x/dt^2 = -k_{\text{eff}}x$.
Refer to the system described above. What is the effective spring constant for the system?
$150\ \text{N/m}$
$100\ \text{N/m}$
$450\ \text{N/m}$
$200\ \text{N/m}$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding spring forces in series configurations through the reciprocal addition rule. When springs are connected in series, they experience the same force but have different displacements that add, resulting in 1/keff = 1/k₁ + 1/k₂. In this problem, two springs with constants k₁ = 300 N/m and k₂ = 150 N/m are connected in series between a wall and a 0.600 kg block. Choice B is correct because 1/keff = 1/300 + 1/150 = 1/300 + 2/300 = 3/300 = 1/100, giving keff = 100 N/m. Choice C (150 N/m) might result from taking the smaller spring constant or averaging incorrectly. To help students: Emphasize that series springs share force but add displacements. Practice the reciprocal formula and recognize that keff is always less than the smallest individual k.
A $0.50\ \text{kg}$ cart on a frictionless horizontal track is attached between two springs: spring 1 on the left with constant $k_1 = 120\ \text{N/m}$ and spring 2 on the right with constant $k_2 = 80\ \text{N/m}$. Each spring is fixed to a wall at its far end, and the cart is connected to both springs so that when the cart is displaced a distance x to the right from equilibrium, the left spring stretches by x and the right spring compresses by x. Both springs obey Hooke’s law, and the restoring forces add. Along the track, the only forces on the cart are the spring forces; weight and normal cancel vertically. For a displacement x, the net restoring force is $F_{\text{net}} = -(k_1 + k_2)x$, consistent with Newton’s second law $\sum F = ma$ for simple harmonic motion.
Given values:
- Mass: m = $0.50\ \text{kg}$
- Spring constants: $k_1 = 120\ \text{N/m}$, $k_2 = 80\ \text{N/m}$
- Motion is horizontal and frictionless
Refer to the system described above. What is the effective spring constant for the system?
$48\ \text{N/m}$
$20\ \text{N/m}$
$200\ \text{N/m}$
$96\ \text{N/m}$
Explanation
This question tests AP Physics C: Mechanics skills, specifically understanding spring forces in parallel configuration and effective spring constants. When springs are connected in parallel (both attached to the same mass), their restoring forces add directly, making the effective spring constant the sum of individual constants. For a rightward displacement x, spring 1 stretches and pulls left with force k₁x, while spring 2 compresses and pushes left with force k₂x, giving total restoring force F = -(k₁ + k₂)x. Choice B is correct because the effective spring constant is keff = k₁ + k₂ = 120 N/m + 80 N/m = 200 N/m. Choice A (48 N/m) is incorrect as it might result from incorrectly treating the springs as in series using the reciprocal formula. To help students: Use free body diagrams to show how forces add in parallel spring systems and contrast with series configurations. Watch for: confusion between parallel and series spring formulas, and sign errors when determining force directions.