Spring Forces
Help Questions
AP Physics C: Mechanics › Spring Forces
A force $$F$$ is required to stretch a single ideal spring with constant $$k$$ by a distance $$x$$. If two identical springs are connected in parallel, what total force is required to stretch the combination by the same distance $$x$$?
$$2F$$
$$F$$
$$F/2$$
$$4F$$
Explanation
For two identical springs of constant $$k$$ in parallel, the equivalent spring constant is $$k_{\text{eq}} = k + k = 2k$$. The original force was $$F = kx$$. The new force $$F'$$ required to stretch the combination by the same distance $$x$$ is $$F' = k_{\text{eq}}x = (2k)x = 2(kx) = 2F$$.
An ideal spring with spring constant $$k$$ is cut into two identical halves. One of the halves is designated Spring A. The two halves are then connected in parallel to form a new spring system, designated System B. What is the ratio of the spring constant of System B to the spring constant of Spring A?
$$2$$
$$4$$
$$1/2$$
$$1$$
Explanation
When a spring is cut in half, its spring constant doubles because stiffness is inversely proportional to length. So, the spring constant of each half (including Spring A) is $$2k$$. When these two halves are connected in parallel to form System B, their equivalent spring constant is the sum: $$k_B = 2k + 2k = 4k$$. The ratio of the spring constant of System B to Spring A is $$k_B / k_A = 4k / (2k) = 2$$.
An ideal spring is fixed at one end to a wall. When a block attached to the other end is at position $$x > 0$$, the spring is stretched. When the block is at $$x < 0$$, the spring is compressed. The equilibrium position is at $$x=0$$. Which statement correctly describes the force $$\vec{F}_s$$ exerted by the spring on the block?
The force $$\vec{F}_s$$ is directed away from the equilibrium position $$x=0$$ regardless of whether the spring is stretched or compressed.
The force $$\vec{F}_s$$ is directed toward $$x=0$$ when the spring is stretched but away from $$x=0$$ when it is compressed.
The force $$\vec{F}_s$$ is directed away from $$x=0$$ when the spring is stretched but toward $$x=0$$ when it is compressed.
The force $$\vec{F}_s$$ is directed toward the equilibrium position $$x=0$$ regardless of whether the spring is stretched or compressed.
Explanation
The force exerted by an ideal spring is a restoring force, meaning it always acts to return the attached object to the equilibrium position. Therefore, if the block is at $$x > 0$$ (stretched), the force is in the negative direction (toward $$x=0$$). If the block is at $$x < 0$$ (compressed), the force is in the positive direction (also toward $$x=0$$).
An ideal spring of natural length $$L_0$$ and spring constant $$k_0$$ is cut into three equal segments. What is the spring constant of one of these segments?
$$k_0\sqrt{3}$$
$$3k_0$$
$$k_0$$
$$k_0/3$$
Explanation
The spring constant is inversely proportional to the natural length of the spring (a shorter spring is stiffer). If the spring is cut into three equal segments, the length of each segment is $$L_0/3$$. The new spring constant $$k'$$ will be three times the original, so $$k' = 3k_0$$.
A non-ideal spring exerts a restoring force on an object given by $$F = -ax - bx^3$$, where $$a$$ and $$b$$ are positive constants. The object is displaced from equilibrium by a small distance $$x_0$$. For this displacement, the contribution from the cubic term is negligible. If the displacement is doubled to $$2x_0$$ such that the cubic term is no longer negligible, how does the magnitude of the restoring force compare to the ideal spring force $$F_{ideal} = k(2x_0)$$ with $$k=a$$?
It cannot be determined without knowing the values of $$a$$ and $$b$$.
It is less than the ideal spring force because the cubic term opposes the linear term.
It is greater than the ideal spring force because the cubic term adds to the magnitude of the linear term.
It is equal to the ideal spring force because the non-ideal effects only matter at very large displacements.
Explanation
The magnitude of the restoring force is $$|F| = | -ax - bx^3 | = ax + bx^3$$ for positive $$x$$. The ideal spring force would be $$F_{ideal} = ax$$. The actual force includes the additional positive term $$bx^3$$. Therefore, for any non-zero displacement, the magnitude of the force from the non-ideal spring is greater than that of an ideal spring with constant $$a$$.
A block is placed on a frictionless horizontal surface between two walls. Two ideal springs, with constants $$k_1$$ and $$k_2$$, are connected to opposite sides of the block and to the walls. The springs are at their natural lengths at the equilibrium position $$x=0$$. If the block is displaced a distance $$x$$ from equilibrium, what is the effective spring constant $$k_{\text{eff}}$$ of the system that provides the net restoring force?
$$|k_1 - k_2|$$
$$\sqrt{k_1^2 + k_2^2}$$
$$k_1 + k_2$$
$$(\frac{1}{k_1} + \frac{1}{k_2})^{-1}$$
Explanation
When the block is displaced by $$x$$, one spring is stretched by $$x$$ and the other is compressed by $$x$$. The stretched spring pulls the block back toward equilibrium with force $$k_1x$$, and the compressed spring pushes the block back toward equilibrium with force $$k_2x$$. Since both forces act in the same restoring direction, the net force is $$F_{\text{net}} = k_1x + k_2x = (k_1 + k_2)x$$. The effective spring constant is therefore $$k_{\text{eff}} = k_1 + k_2$$.
A student has two ideal springs, A and B, with spring constants satisfying $$k_A > k_B$$. The student connects them in series, hangs a block from the combination, and measures the stretch. The student then connects them in parallel, hangs the same block, and measures the stretch. The student claims that the stretch will be greater in the series configuration. Which of the following correctly evaluates this claim?
The claim is correct because the equivalent spring constant for the series combination is smaller than for the parallel combination, resulting in a larger stretch for the same force.
The claim is incorrect because the series combination results in a stiffer spring system, which means it will stretch less than the more flexible parallel combination.
The claim is correct because in series, the block's weight is fully applied to each spring sequentially, whereas in parallel the weight is shared between the springs.
The claim is incorrect because the parallel combination is a stronger overall spring system, but this means it is harder to stretch, resulting in less stretch.
Explanation
The equivalent spring constant in series is $$k_s = (1/k_A + 1/k_B)^{-1}$$, which is always less than both $$k_A$$ and $$k_B$$. The equivalent constant in parallel is $$k_p = k_A + k_B$$, which is always greater than both. Since stretch $$x = F/k$$ and the force (the block's weight) is the same, the smaller spring constant (series) will result in a greater stretch. The student's claim is correct.
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?
$200\ \text{N/m}$
$96\ \text{N/m}$
$48\ \text{N/m}$
$150\ \text{N/m}$
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?
$100\ \text{N/m}$
$200\ \text{N/m}$
$900\ \text{N/m}$
$450\ \text{N/m}$
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?
$150\ \text{N/m}$
$450\ \text{N/m}$
$200\ \text{N/m}$
$100\ \text{N/m}$
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.