AP Physics C Mechanics Quiz: Spring Forces
20 questions · exam conditions
0:00
Spring ForcesQuestion 1 of 20

A force FF is required to stretch a single ideal spring with constant kk by a distance xx. If two identical springs are connected in parallel, what total force is required to stretch the combination by the same distance xx?

F/2F/2
FF
2F2F
4F4F
← Back to quizzes

AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Spring Forces

Practice Spring Forces in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Spring Forces, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A force FF is required to stretch a single ideal spring with constant kk by a distance xx. If two identical springs are connected in parallel, what total force is required to stretch the combination by the same distance xx?

  1. F/2F/2
  2. FF
  3. 2F2F (correct answer)
  4. 4F4F
Explanation: For two identical springs of constant kk in parallel, the equivalent spring constant is keq=k+k=2kk_{\text{eq}} = k + k = 2k. The original force was F=kxF = kx. The new force FF' required to stretch the combination by the same distance xx is F=keqx=(2k)x=2(kx)=2FF' = k_{\text{eq}}x = (2k)x = 2(kx) = 2F.

Question 2

An ideal spring with spring constant kk 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?

  1. 1/21/2
  2. 11
  3. 22 (correct answer)
  4. 44
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 2k2k. When these two halves are connected in parallel to form System B, their equivalent spring constant is the sum: kB=2k+2k=4kk_B = 2k + 2k = 4k. The ratio of the spring constant of System B to Spring A is kB/kA=4k/(2k)=2k_B / k_A = 4k / (2k) = 2.

Question 3

An ideal spring is fixed at one end to a wall. When a block attached to the other end is at position x>0x > 0, the spring is stretched. When the block is at x<0x < 0, the spring is compressed. The equilibrium position is at x=0x=0. Which statement correctly describes the force Fs\vec{F}_s exerted by the spring on the block?

  1. The force Fs\vec{F}_s is directed toward the equilibrium position x=0x=0 regardless of whether the spring is stretched or compressed. (correct answer)
  2. The force Fs\vec{F}_s is directed away from the equilibrium position x=0x=0 regardless of whether the spring is stretched or compressed.
  3. The force Fs\vec{F}_s is directed toward x=0x=0 when the spring is stretched but away from x=0x=0 when it is compressed.
  4. The force Fs\vec{F}_s is directed away from x=0x=0 when the spring is stretched but toward x=0x=0 when it is 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>0x > 0 (stretched), the force is in the negative direction (toward x=0x=0). If the block is at x<0x < 0 (compressed), the force is in the positive direction (also toward x=0x=0).

Question 4

A non-ideal spring exerts a restoring force on an object given by F=axbx3F = -ax - bx^3, where aa and bb are positive constants. The object is displaced from equilibrium by a small distance x0x_0. For this displacement, the contribution from the cubic term is negligible. If the displacement is doubled to 2x02x_0 such that the cubic term is no longer negligible, how does the magnitude of the restoring force compare to the ideal spring force Fideal=k(2x0)F_{ideal} = k(2x_0) with k=ak=a?

  1. It is less than the ideal spring force because the cubic term opposes the linear term.
  2. It is equal to the ideal spring force because the non-ideal effects only matter at very large displacements.
  3. It is greater than the ideal spring force because the cubic term adds to the magnitude of the linear term. (correct answer)
  4. It cannot be determined without knowing the values of aa and bb.
Explanation: The magnitude of the restoring force is F=axbx3=ax+bx3|F| = | -ax - bx^3 | = ax + bx^3 for positive xx. The ideal spring force would be Fideal=axF_{ideal} = ax. The actual force includes the additional positive term bx3bx^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 aa.

Question 5

A block is placed on a frictionless horizontal surface between two walls. Two ideal springs, with constants k1k_1 and k2k_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=0x=0. If the block is displaced a distance xx from equilibrium, what is the effective spring constant keffk_{\text{eff}} of the system that provides the net restoring force?

  1. k1+k2k_1 + k_2 (correct answer)
  2. (1k1+1k2)1(\frac{1}{k_1} + \frac{1}{k_2})^{-1}
  3. k1k2|k_1 - k_2|
  4. k12+k22\sqrt{k_1^2 + k_2^2}
Explanation: When the block is displaced by xx, one spring is stretched by xx and the other is compressed by xx. The stretched spring pulls the block back toward equilibrium with force k1xk_1x, and the compressed spring pushes the block back toward equilibrium with force k2xk_2x. Since both forces act in the same restoring direction, the net force is Fnet=k1x+k2x=(k1+k2)xF_{\text{net}} = k_1x + k_2x = (k_1 + k_2)x. The effective spring constant is therefore keff=k1+k2k_{\text{eff}} = k_1 + k_2.

Question 6

A student has two ideal springs, A and B, with spring constants satisfying kA>kBk_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?

  1. 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. (correct answer)
  2. 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.
  3. 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.
  4. 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.
Explanation: The equivalent spring constant in series is ks=(1/kA+1/kB)1k_s = (1/k_A + 1/k_B)^{-1}, which is always less than both kAk_A and kBk_B. The equivalent constant in parallel is kp=kA+kBk_p = k_A + k_B, which is always greater than both. Since stretch x=F/kx = 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.

Question 7

A 0.250 kg0.250\ \text{kg} block is attached to two ideal springs on a horizontal frictionless table. Spring 1 has constant k1=120 N/mk_1 = 120\ \text{N/m} and Spring 2 has constant k2=80.0 N/mk_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:

  • k1=120 N/mk_1 = 120\ \text{N/m}
  • k2=80.0 N/mk_2 = 80.0\ \text{N/m}
  • m=0.250 kgm = 0.250\ \text{kg}

Forces and model: For a displacement x, each spring exerts a restoring force Fs1=k1xF_{s1} = -k_1 x and Fs2=k2xF_{s2} = -k_2 x. The net restoring force is Fnet=(k1+k2)x,F_{\text{net}} = -(k_1 + k_2)x, so Newton's Second Law becomes md2xdt2=(k1+k2)x.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?

  1. 48 N/m48\ \text{N/m}
  2. 200 N/m200\ \text{N/m} (correct answer)
  3. 96 N/m96\ \text{N/m}
  4. 150 N/m150\ \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.

Question 8

A 2.0 kg2.0\ \text{kg} cart is attached to two springs connected in series along a horizontal frictionless track. Spring A has constant kA=300 N/mk_A = 300\ \text{N/m} and spring B has constant kB=600 N/mk_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 1keff=1kA+1kB\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 kg2.0\ \text{kg}
  • Spring constants: kA=300 N/mk_A = 300\ \text{N/m}, kB=600 N/mk_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?

  1. 900 N/m900\ \text{N/m}
  2. 200 N/m200\ \text{N/m} (correct answer)
  3. 450 N/m450\ \text{N/m}
  4. 100 N/m100\ \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.

Question 9

A 0.600 kg0.600\ \text{kg} block is connected to two ideal springs in series on a frictionless horizontal surface. Spring 1 has constant k1=300 N/mk_1 = 300\ \text{N/m} and Spring 2 has constant k2=150 N/mk_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:

  • k1=300 N/mk_1 = 300\ \text{N/m}
  • k2=150 N/mk_2 = 150\ \text{N/m}
  • m=0.600 kgm = 0.600\ \text{kg}

Forces and model: In series, both springs carry the same force magnitude F while their extensions add: x=x1+x2x = x_1 + x_2. Hooke's Law gives F=k1x1=k2x2F = k_1 x_1 = k_2 x_2, so the equivalent spring constant satisfies 1keff=1k1+1k2.\frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2}. The motion then satisfies md2x/dt2=keffxm\,d^2x/dt^2 = -k_{\text{eff}}x.

Refer to the system described above. What is the effective spring constant for the system?

  1. 450 N/m450\ \text{N/m}
  2. 100 N/m100\ \text{N/m} (correct answer)
  3. 150 N/m150\ \text{N/m}
  4. 200 N/m200\ \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.

Question 10

A 0.50 kg0.50\ \text{kg} cart on a frictionless horizontal track is attached between two springs: spring 1 on the left with constant k1=120 N/mk_1 = 120\ \text{N/m} and spring 2 on the right with constant k2=80 N/mk_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 Fnet=(k1+k2)xF_{\text{net}} = -(k_1 + k_2)x, consistent with Newton's second law F=ma\sum F = ma for simple harmonic motion.

Given values:

  • Mass: m = 0.50 kg0.50\ \text{kg}
  • Spring constants: k1=120 N/mk_1 = 120\ \text{N/m}, k2=80 N/mk_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?

  1. 48 N/m48\ \text{N/m}
  2. 200 N/m200\ \text{N/m} (correct answer)
  3. 96 N/m96\ \text{N/m}
  4. 20 N/m20\ \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.

Question 11

An ideal horizontal spring with a spring constant of k=50N/mk = 50 \, \text{N/m} is attached to a block on a frictionless surface. The block is pulled, stretching the spring by 0.20m0.20 \, \text{m} from its equilibrium position. What is the magnitude of the restoring force exerted by the spring on the block?

  1. 1.0N1.0 \, \text{N}
  2. 10N10 \, \text{N} (correct answer)
  3. 25N25 \, \text{N}
  4. 250N250 \, \text{N}
Explanation: According to Hooke's Law, the magnitude of the restoring force exerted by an ideal spring is given by F=kxF = kx, where kk is the spring constant and xx is the displacement from equilibrium. Substituting the given values: F=(50N/m)(0.20m)=10NF = (50 \, \text{N/m})(0.20 \, \text{m}) = 10 \, \text{N}.

Question 12

A block of mass m=2.0kgm = 2.0 \, \text{kg} is suspended from a vertical ideal spring, causing the spring to stretch by 0.10m0.10 \, \text{m} from its natural length once it reaches equilibrium. What is the spring constant kk of the spring? (Use g10m/s2g \approx 10 \, \text{m/s}^2)

  1. 0.02N/m0.02 \, \text{N/m}
  2. 0.5N/m0.5 \, \text{N/m}
  3. 20N/m20 \, \text{N/m}
  4. 200N/m200 \, \text{N/m} (correct answer)
Explanation: In equilibrium, the upward spring force FsF_s balances the downward gravitational force FgF_g. So, Fs=FgF_s = F_g, which means kx=mgkx = mg. Solving for the spring constant kk gives k=mg/x=(2.0kg)(10m/s2)/(0.10m)=200N/mk = mg/x = (2.0 \, \text{kg})(10 \, \text{m/s}^2) / (0.10 \, \text{m}) = 200 \, \text{N/m}.

Question 13

Two ideal springs with spring constants k1=100N/mk_1 = 100 \, \text{N/m} and k2=300N/mk_2 = 300 \, \text{N/m} are attached in parallel to a block on a frictionless horizontal surface. What is the equivalent spring constant keqk_{\text{eq}} of this spring system?

  1. 75N/m75 \, \text{N/m}
  2. 200N/m200 \, \text{N/m}
  3. 400N/m400 \, \text{N/m} (correct answer)
  4. 500N/m500 \, \text{N/m}
Explanation: For springs connected in parallel, the equivalent spring constant is the sum of the individual spring constants: keq=k1+k2k_{\text{eq}} = k_1 + k_2. Substituting the given values, keq=100N/m+300N/m=400N/mk_{\text{eq}} = 100 \, \text{N/m} + 300 \, \text{N/m} = 400 \, \text{N/m}.

Question 14

Two ideal springs with spring constants k1=100N/mk_1 = 100 \, \text{N/m} and k2=300N/mk_2 = 300 \, \text{N/m} are connected in series. What is the equivalent spring constant keqk_{\text{eq}} of this combination?

  1. 75N/m75 \, \text{N/m} (correct answer)
  2. 200N/m200 \, \text{N/m}
  3. 400N/m400 \, \text{N/m}
  4. 500N/m500 \, \text{N/m}
Explanation: For springs connected in series, the reciprocal of the equivalent spring constant is the sum of the reciprocals of the individual spring constants: 1/keq=1/k1+1/k21/k_{\text{eq}} = 1/k_1 + 1/k_2. Substituting the values: 1/keq=1/100+1/300=(3+1)/300=4/3001/k_{\text{eq}} = 1/100 + 1/300 = (3+1)/300 = 4/300. Therefore, keq=300/4=75N/mk_{\text{eq}} = 300/4 = 75 \, \text{N/m}.

Question 15

A block of mass MM rests on a frictionless plane inclined at an angle θ\theta to the horizontal. The block is attached to an ideal spring with spring constant kk, which is fixed to the top of the incline. The spring is stretched by a distance xx from its natural length to hold the block in equilibrium. What is the expression for xx?

  1. Mg/kMg/k
  2. Mgsinθ/kMg\sin\theta / k (correct answer)
  3. Mgcosθ/kMg\cos\theta / k
  4. k/(Mgsinθ)k / (Mg\sin\theta)
Explanation: In equilibrium, the net force on the block is zero. The forces acting along the incline are the spring force Fs=kxF_s = kx (up the incline) and the component of gravity parallel to the incline Fg,=MgsinθF_{g, \parallel} = Mg\sin\theta (down the incline). Setting these equal gives kx=Mgsinθkx = Mg\sin\theta. Solving for xx yields x=Mgsinθ/kx = Mg\sin\theta / k.

Question 16

A block of mass mm is attached to a horizontal ideal spring of constant kk. The block is displaced a distance AA from its equilibrium position x=0x=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?

  1. 00
  2. kA/mkA/m (correct answer)
  3. m/(kA)m/(kA)
  4. mg/kmg/k
Explanation: At the moment of release, the only horizontal force acting on the block is the spring force, which has a magnitude of Fs=kAF_s = kA. According to Newton's second law, Fnet=maF_{net} = ma. Therefore, kA=makA = ma. Solving for the acceleration aa gives a=kA/ma = kA/m.

Question 17

The restoring force exerted by an ideal spring has magnitude FF when it is stretched a distance xx from its equilibrium position. If the spring is instead stretched by a distance 3x3x, what is the new magnitude of the restoring force?

  1. F/3F/3
  2. FF
  3. 3F3F (correct answer)
  4. 9F9F
Explanation: For an ideal spring, the restoring force is directly proportional to the displacement from equilibrium (F=kxF = kx). If the initial force is F=kxF = kx, the new force FF' for a displacement of 3x3x will be F=k(3x)=3(kx)=3FF' = k(3x) = 3(kx) = 3F.

Question 18

A mass mm is hung from an ideal spring of constant kk, causing it to stretch a distance xx. A second, identical spring is then attached in series with the first, and the same mass mm is hung from the combination. What is the total stretch of the two-spring combination?

  1. x/2x/2
  2. xx
  3. x2x\sqrt{2}
  4. 2x2x (correct answer)
Explanation: For two identical springs of constant kk in series, the equivalent spring constant is keq=(1/k+1/k)1=(2/k)1=k/2k_{\text{eq}} = (1/k + 1/k)^{-1} = (2/k)^{-1} = k/2. The original stretch was x=mg/kx = mg/k. The new stretch xx' is x=mg/keq=mg/(k/2)=2(mg/k)=2xx' = mg/k_{\text{eq}} = mg/(k/2) = 2(mg/k) = 2x.

Question 19

The magnitude of the force FF required to stretch an ideal spring is measured as a function of the spring's elongation xx. A graph of FF versus xx is created, which is a straight line passing through the origin. The slope of this line represents which physical quantity?

  1. The work done in stretching the spring.
  2. The spring constant kk. (correct answer)
  3. The reciprocal of the spring constant, 1/k1/k.
  4. The acceleration of a mass attached to the spring.
Explanation: Hooke's Law is F=kxF = kx. When plotting FF on the y-axis and xx on the x-axis, this equation takes the form y=mx+by = mx+b, where the slope mm is the spring constant kk and the y-intercept bb is zero. Thus, the slope of an F-vs-x graph for an ideal spring is the spring constant.

Question 20

An ideal spring with spring constant kk is initially compressed by a distance x0x_0 from its equilibrium length. An external force is applied to compress it further to a final compression of 3x03x_0. What is the change in the magnitude of the force exerted by the spring?

  1. kx0kx_0
  2. 2kx02kx_0 (correct answer)
  3. 3kx03kx_0
  4. 4kx04kx_0
Explanation: The magnitude of the spring force is given by F=kxF = kx, where xx is the magnitude of the displacement from equilibrium. The initial force magnitude is Fi=kx0F_i = kx_0. The final force magnitude is Ff=k(3x0)=3kx0F_f = k(3x_0) = 3kx_0. The change in the magnitude of the force is ΔF=FfFi=3kx0kx0=2kx0\Delta F = F_f - F_i = 3kx_0 - kx_0 = 2kx_0.