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AP Physics 1: Algebra-Based Practice Quiz: Spring Forces

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

According to the provided description of an ideal spring, the force it exerts is directly proportional to what quantity?

All Questions (10)

According to the provided description of an ideal spring, the force it exerts is directly proportional to what quantity?

A) The change in its length from its relaxed position.

B) The total length of the spring.

C) The mass attached to the spring.

D) The acceleration of the object attached to the spring.

Correct Answer: A

Content 2 states that an ideal spring "exerts a force that is proportional to the change in its length as measured from its relaxed length." This change in length is represented as Δx in Hooke's Law.

An object attached to an ideal spring is pulled to the right, stretching the spring away from its equilibrium position. In which direction does the spring exert a force on the object?

A) To the right, in the direction of the displacement.

B) To the left, toward the equilibrium position.

C) Upward, perpendicular to the displacement.

D) The direction cannot be determined without knowing the spring constant.

Correct Answer: B

Content 4 states, "The force exerted on an object by a spring is always directed toward the equilibrium position of the object-spring system." Since the object was displaced to the right, the equilibrium position is to the left, so the force is directed to the left.

In Hooke's Law, $\vec{F}_s = -k \Delta \vec{x}$, what is the physical significance of the negative sign?

A) It indicates that the spring force is a non-conservative force.

B) It signifies that the spring constant, k, is always a negative value.

C) It shows that the spring force vector is always directed opposite to the displacement vector.

D) It implies that the force exerted by the spring decreases as it is stretched.

Correct Answer: C

The negative sign indicates that the force vector ($\vec{F}_s$) points in the opposite direction of the displacement vector ($\Delta \vec{x}$). This means the force is a restoring force, always acting to bring the object back toward the equilibrium position, as stated in Content 4.

An ideal spring is stretched a distance Δx from its equilibrium position, causing it to exert a restoring force of magnitude F. If the spring is instead stretched a distance of 2Δx, what is the new magnitude of the restoring force?

A) F/2

B) F

C) 2F

D) 4F

Correct Answer: C

Content 2 states the force is proportional to the change in length. Hooke's Law (Content 3) gives the magnitude as Fs = kΔx. Since the force is directly proportional to the displacement Δx, doubling the displacement will double the magnitude of the force.

Which of the following is a defining characteristic of an "ideal spring" according to the provided text?

A) It has a significant mass.

B) The force it exerts is constant regardless of displacement.

C) It has negligible mass.

D) It can only be stretched, not compressed.

Correct Answer: C

Content 2 explicitly states, "An ideal spring has negligible mass..."

At which point does an ideal spring exert zero force on an attached object?

A) At the point of maximum compression.

B) At the point of maximum stretch.

C) When it is at its relaxed length.

D) An ideal spring always exerts a non-zero force.

Correct Answer: C

The force is proportional to the change in length from its relaxed length (Δx). When the spring is at its relaxed length, the change in length is zero (Δx = 0). According to Hooke's Law, Fs = -k(0), the force is zero. This position is the equilibrium position.

An object-spring system is at equilibrium. The object is then pushed to the left, compressing the spring. What is the direction of the force exerted by the spring on the object?

A) To the left, in the direction of the displacement.

B) Downward, perpendicular to the displacement.

C) To the right, toward the equilibrium position.

D) The force is zero because the spring is compressed.

Correct Answer: C

According to Content 4, the force is always directed toward the equilibrium position. If the spring is compressed by pushing the object to the left, the equilibrium position is to the right. Therefore, the spring pushes the object to the right.

In the Hooke's Law equation, $\vec{F}_s = -k \Delta \vec{x}$, the term $\Delta \vec{x}$ precisely represents the:

A) total length of the spring.

B) vector for the change in length from the spring's relaxed position.

C) spring constant, or stiffness.

D) final position of the object.

Correct Answer: B

Content 2 describes the force as proportional to "the change in its length as measured from its relaxed length." Content 3 formalizes this as the vector $\Delta \vec{x}$ in Hooke's Law, which represents the displacement from the equilibrium (relaxed) position.

If a graph is created with the magnitude of the spring force, Fs, on the vertical axis and the magnitude of the displacement from equilibrium, |Δx|, on the horizontal axis, what would the graph look like for an ideal spring?

A) A horizontal line, indicating constant force.

B) A parabola opening upwards.

C) A straight line passing through the origin.

D) A curve that approaches a maximum value.

Correct Answer: C

Hooke's Law for magnitude is Fs = k|Δx|. This is a linear relationship in the form y = mx, where Fs is y, |Δx| is x, and the spring constant k is the constant slope m. This relationship is described as a straight line passing through the origin, consistent with the force being "proportional to the change in its length."

In Hooke's Law, the spring constant 'k' is a measure of the stiffness of the spring. A spring with a larger value of 'k' would:

A) be easier to stretch.

B) exert a larger force for the same displacement.

C) have a smaller mass.

D) exert a smaller force for the same displacement.

Correct Answer: B

From the equation Fs = kΔx (magnitude), the force Fs is directly proportional to k. If k is larger, then for the same displacement Δx, the resulting force Fs will be larger. This means the spring is stiffer and requires more force to stretch or compress.