Spring Forces - AP Physics C Mechanics

Spring X has a spring constant $k$ . Spring Y has a spring constant $2k$ . Both springs are stretched by the same constant force $F$ . If spring X stretches by a distance $x$ , by what distance does spring Y stretch?
Answer: $x/2$ From Hooke's Law, the stretch distance is $d = F/k$ . For spring X, $x = F/k$ . For spring Y, its stretch distance $y$ is $y = F/(2k)$ . We can substitute $F=kx$ into the second equation: $y = (kx)/(2k) = x/2$ . Spring Y, being stiffer, stretches half as much for the same force.
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?
Answer: $100\ \text{N/m}$ 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 spring with natural length $L_0$ hangs vertically. When a mass $m$ is attached, the spring stretches to a new equilibrium length. If the mass is then displaced downward by a small amount $y$ and released, it oscillates with period $T$ . Which expression correctly represents the spring constant $k$ ?
Answer: $k = \frac{4\pi^2 m}{T^2}$ For a mass on a vertical spring oscillating about its equilibrium position, the period is $T = 2\pi\sqrt{\frac{m}{k}}$ , which gives $k = \frac{4\pi^2 m}{T^2}$ . The fact that the spring hangs vertically doesn't change the oscillation period formula - gravity only shifts the equilibrium position. Choice B incorrectly includes gravity in the spring constant formula. Choice C inverts the relationship. Choice D confuses static equilibrium with dynamic oscillation.
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?
Answer: $200\ \text{N/m}$ 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.
Two ideal springs with spring constants $k_1 = 100 \, \text{N/m}$ and $k_2 = 300 \, \text{N/m}$ are connected in series. What is the equivalent spring constant $k_{\text{eq}}$ of this combination?
Answer: $75 \, \text{N/m}$ For springs connected in series, the reciprocal of the equivalent spring constant is the sum of the reciprocals of the individual spring constants: $1/k_{\text{eq}} = 1/k_1 + 1/k_2$ . Substituting the values: $1/k_{\text{eq}} = 1/100 + 1/300 = (3+1)/300 = 4/300$ . Therefore, $k_{\text{eq}} = 300/4 = 75 \, \text{N/m}$ .
A spring-mass system oscillates horizontally with amplitude $A = 0.12 \text{ m}$ and frequency $f = 2.0 \text{ Hz}$ . At $t = 0$ , the mass is at maximum displacement in the positive direction. What is the first time after $t = 0$ when the spring force equals zero?
Answer: $0.125 \text{ s}$ The spring force equals zero when the mass passes through equilibrium ( $x = 0$ ). With $\omega = 2\pi f = 4\pi \text{ rad/s}$ and initial conditions $x(0) = A$ , the position is $x(t) = A\cos(\omega t) = 0.12\cos(4\pi t)$ . The force is zero when $x = 0$ , so $\cos(4\pi t) = 0$ . The first time this occurs is when $4\pi t = \frac{\pi}{2}$ , giving $t = \frac{1}{8} = 0.125 \text{ s}$ . Choice B uses $\sin$ instead of $\cos$ . Choice C gives the half-period. Choice D gives the full period.
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?
Answer: $k_1 + k_2$ 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 block of mass $M$ rests on a frictionless plane inclined at an angle $\theta$ to the horizontal. The block is attached to an ideal spring with spring constant $k$ , which is fixed to the top of the incline. The spring is stretched by a distance $x$ from its natural length to hold the block in equilibrium. What is the expression for $x$ ?
Answer: $Mg\sin\theta / k$ In equilibrium, the net force on the block is zero. The forces acting along the incline are the spring force $F_s = kx$ (up the incline) and the component of gravity parallel to the incline $F_{g, \parallel} = Mg\sin\theta$ (down the incline). Setting these equal gives $kx = Mg\sin\theta$ . Solving for $x$ yields $x = Mg\sin\theta / k$ .
A block of mass $m$ is attached to a horizontal ideal spring of constant $k$ . The block is displaced a distance $A$ from its equilibrium position $x=0$ and released from rest on a frictionless surface. What is the magnitude of the initial acceleration of the block at the moment of release?
Answer: $kA/m$ At the moment of release, the only horizontal force acting on the block is the spring force, which has a magnitude of $F_s = kA$ . According to Newton's second law, $F_{net} = ma$ . Therefore, $kA = ma$ . Solving for the acceleration $a$ gives $a = kA/m$ .
A mass $m$ is hung from an ideal spring of constant $k$ , causing it to stretch a distance $x$ . A second, identical spring is then attached in series with the first, and the same mass $m$ is hung from the combination. What is the total stretch of the two-spring combination?
Answer: $2x$ For two identical springs of constant $k$ in series, the equivalent spring constant is $k_{\text{eq}} = (1/k + 1/k)^{-1} = (2/k)^{-1} = k/2$ . The original stretch was $x = mg/k$ . The new stretch $x'$ is $x' = mg/k_{\text{eq}} = mg/(k/2) = 2(mg/k) = 2x$ .
A mass $m$ oscillates on a horizontal spring with amplitude $A$ and angular frequency $\omega$ . At what displacement from equilibrium is the magnitude of the acceleration equal to half the maximum acceleration?
Answer: $\frac{A}{2}$ For simple harmonic motion, $a = -\omega^2 x$ . Maximum acceleration occurs at maximum displacement: $a_{max} = \omega^2 A$ . We want $|a| = \frac{1}{2}a_{max}$ , so $\omega^2|x| = \frac{1}{2}\omega^2 A$ . Therefore $|x| = \frac{A}{2}$ . Choice A gives one-quarter maximum acceleration. Choice C gives $\frac{1}{\sqrt{2}}$ times maximum acceleration. Choice D gives $\frac{\sqrt{3}}{2}$ times maximum acceleration.
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 $k=a$ ?
Answer: It is greater than the ideal spring force because the cubic term adds to the magnitude of the linear term. 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 light, rigid rod is suspended horizontally by two identical vertical ideal springs, each with spring constant $k$ , attached to its ends. A block of mass $M$ is then placed on the center of the rod. By what vertical distance does the rod move downward to reach its new equilibrium position?
Answer: $Mg/(2k)$ Since the springs are supporting the rigid rod in parallel, their effective spring constant is $k_{\text{eq}} = k + k = 2k$ . At the new equilibrium position, the total upward spring force must balance the downward gravitational force of the block. Let $y$ be the downward displacement. Then $F_{\text{eq}} = k_{\text{eq}}y = (2k)y$ . Setting this equal to the weight, $(2k)y = Mg$ . Solving for $y$ gives $y = Mg/(2k)$ .
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?
Answer: 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 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 block of mass $m = 2.0 \, \text{kg}$ is suspended from a vertical ideal spring, causing the spring to stretch by $0.10 \, \text{m}$ from its natural length once it reaches equilibrium. What is the spring constant $k$ of the spring? (Use $g \approx 10 \, \text{m/s}^2$ )
Answer: $200 \, \text{N/m}$ In equilibrium, the upward spring force $F_s$ balances the downward gravitational force $F_g$ . So, $F_s = F_g$ , which means $kx = mg$ . Solving for the spring constant $k$ gives $k = mg/x = (2.0 \, \text{kg})(10 \, \text{m/s}^2) / (0.10 \, \text{m}) = 200 \, \text{N/m}$ .
The magnitude of the force $F$ required to stretch an ideal spring is measured as a function of the spring's elongation $x$ . A graph of $F$ versus $x$ is created, which is a straight line passing through the origin. The slope of this line represents which physical quantity?
Answer: The spring constant $k$ . Hooke's Law is $F = kx$ . When plotting $F$ on the y-axis and $x$ on the x-axis, this equation takes the form $y = mx+b$ , where the slope $m$ is the spring constant $k$ and the y-intercept $b$ is zero. Thus, the slope of an F-vs-x graph for an ideal spring is the spring constant.
Two identical ideal springs, each with spring constant $k$ , are attached to a block of mass $m$ and to fixed supports. One spring is horizontal, and the other is vertical, such that they pull on the block at a $90^{\circ}$ angle to each other. The block is on a frictionless horizontal surface and the vertical spring supports it against gravity. The block is displaced horizontally by a distance $x$ from equilibrium. What is the magnitude of the net restoring force on the block?
Answer: $kx$ The vertical spring's force only acts in the vertical direction to counteract gravity, and since the displacement is purely horizontal, the vertical forces remain in equilibrium. The only force contributing to the horizontal restoration is the horizontal spring. Therefore, the net restoring force is simply the force from the horizontal spring, which is $F = kx$ .
A $1.5 \text{ kg}$ block on a frictionless horizontal surface is attached to a spring with $k = 240 \text{ N/m}$ . The block is displaced $0.25 \text{ m}$ from equilibrium and released. At the instant when the block is $0.15 \text{ m}$ from equilibrium, what is the magnitude of its acceleration?
Answer: $24 \text{ m/s}^2$ When you encounter a spring-mass system problem, you're dealing with simple harmonic motion where the restoring force follows Hooke's Law: $F = -kx$ . The key insight is that acceleration depends only on the displacement from equilibrium, not on how the system got there. At any position $x$ from equilibrium, the spring force is $F = kx$ (taking magnitude), and using Newton's second law: $F = ma$ , so $ma = kx$ . Therefore: $a = \frac{kx}{m}$ . When the block is $0.15 \text{ m}$ from equilibrium: $a = \frac{kx}{m} = \frac{(240 \text{ N/m})(0.15 \text{ m})}{1.5 \text{ kg}} = \frac{36}{1.5} = 24 \text{ m/s}^2$ This confirms answer C is correct. Let's examine why the other answers are wrong: A) 12 m/s² likely comes from using the wrong displacement value or making an arithmetic error in the calculation. B) 30 m/s² might result from incorrectly using the initial displacement ( $0.25 \text{ m}$ ) instead of the current position, or from a calculation error. D) 40 m/s² could come from forgetting to account for the mass properly or making an error in applying Hooke's Law. Strategy tip: In spring problems, always identify what displacement you need (current position, not initial conditions), then apply $a = \frac{kx}{m}$ directly. The initial displacement of $0.25 \text{ m}$ is given to potentially mislead you—focus on where the block is now, not where it started.
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?
Answer: $200\ \text{N/m}$ 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.
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?
Answer: $3k_0$ 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 $k' = 3k_0$ .

Start this quiz