# Hooke’s law and elasticity – problems and solutions

Hooke’s law and elasticity – problems and solutions

The change in length

1. A rod has a length of L, pulled by a force of F. The amount of elongation is ∆L. What is the magnitude of the force if the change in length is 4∆L.

Known :

Force 1 (F1) = F

The change in length 1 (∆L1) = ∆L

The change in length 2 (∆L2) = 4 ∆L

Wanted : Force 2 (F2)

Solution :

The equation of Hooke’s law

k = F / ΔL

k = constant of elasticity, F = force of F, ΔL = the change in length

k1 = k2

F1 / ∆L1 = F2 / ∆L2

F / ΔL = F2 / 4ΔL

F / 1 = F2 / 4

F = F2 / 4

F2 = 4F

2. There springs are connected in series-parallel, as shown in figure below. Spring 1 has constant 200 N/m, spring 2 has constant 200 N/m and spring 3 has constant 200 N/m. The mass of object is 100 gram and acceleration due to gravity is 10 m/s2. What is the change in length of the equivalent spring.

Known :

Object’s mass (m) = 100 gram = 0.1 kg

k1 = k2 = k3 = 200 N/m

w = m g = (0.1 kg)(10 m/s2) = 1 kg m/s2 = 1 Newton

Wanted : The change in length of the equivalent spring.

Solution :

Determine the equivalent spring constant :

Spring 2 (k2) and spring 3 (k3) are connected in parallel. The equivalent spring constant :

kp = k2 + k3 = 200 + 200 = 400 Nm−1

Spring 1 (k1) and spring p (kP) are connected in series. The equivalent spring constant :

1/ks = 1/kp + 1/k1 = 1/400 + 1/200 = 1/400 + 2/400 = 3/400

ks = 400/3 Nm−1

The equivalent spring constant is 400/3 Nm-1

Determine the change in length of the equivalent spring :

The equation of Hooke’s law :

∆x = F / k = w / k

The change in length of the equivalent spring :

∆x = w / k

∆x = 1 : 400/3 = 1 x 3/400 = 3/400 = 0.0075 m = 0.75 cm

The constant of spring

3. What is the constant of spring according to the data in the table below.

Solution :

The equation of Hooke’s law :

k = F / Δx

Constant of spring :

k = 0.98 / 0.0008 = 1.96 / 0.0016 = 2.94 / 0.0024 = 3.92 / 0.0032 = 1.225 N/m

4. Three springs are connected in series-parallel as shown in figure below. The constant of spring k1 = k2 = 3 Nm−1 and k3 = 6 Nm−1. What is the constant of the equivalent of spring.

Known :

Constant of spring 1 (k1) = constant of spring 2 (k2) = 3 Nm−1

Constant of spring 3 (k3) = 6 Nm−1

Wanted : constant of the equivalent spring (k)

Solution :

Spring 1 (k1) and spring 2 (k2) are connected in parallel. The Constant of the equivalent spring :

kp = k1 + k2 = 3 + 3 = 6 Nm−1

Spring p (kP) and spring 3 (k3 ) are connected in series. The constant of the equivalent spring :

1/ks = 1/kp + 1/k 3 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6

ks = 6/2 = 3 Nm−1

The constant of the equivalent of spring = 3 Nm−1.

5. A spring with length of L, pulled by weight of w. According to data in table below, what is the constant of the equivalent spring :

Solution :

k = F / Δx

Constant of spring :

k = 10 / 0.02 = 20 / 0.04 = 30 / 0.06 = 40 / 0.08 = 500 N/m

6. According to data in table below, what is the constant of the equivalent spring :

Solution :

k = F / Δx = w / Δx = m g / Δx

k = constant of elasticity, w = weight, m = mass, g = acceleration due to gravity, Δx = the change in length

Spring constant :

k = 2 / 0.05 = 4 / 0.1 = 6 / 0.15 = 8 / 0.20 = 10 / 0.25 = 40 N/m

7. If k1 = 4k, what is the constant of the equivalent spring.

Solution :

Two springs are connected in parallel. The constant of the equivalent spring :

kp = k + k = 2k

Two springs are connected in series. The constant of the equivalent spring

1/ks = 1/kp + 1/k1 = 1 / 2k + 1 / 4k = 2 / 4k + 1 / 4k = 3 / 4k

ks = 4k/3

8. According to data in table below, what is the constant of the equivalent spring :

Solution :

The equation of Hooke’s law :

k = F / ΔL

Constant of spring :

k = 2 / 0.0050 = 3 / 0.0075 = 4 / 0.01 = 400 Nm-1

9. The smallest constant is…

Solution

The equation of Hooke’s law :

k = F / Δx

k = constant of elasticity, F = force, Δx = the change in length

Constant of elasticity :

kA = F / Δx = 1 / 0.05 = 20 N/m

kB = F / Δx = 2 / 0.025 = 80 N/m

kC = F / Δx = 1 / 0.025 = 40 N/m

kD = F / Δx = 2 / 0.05 = 40 N/m

kE = F / Δx = 2 / 0.25 = 8 N/m

10. What is the largest constant according to data in table below.

Solution :

The equation of the Hooke’s law :

k = F / Δx

kA = 7 / 0.035 = 200 Nm-1

kB = 8 / 0.025 = 320 Nm-1

kC = 6 / 0.020 = 300 Nm-1

kD = 9 / 0.045 = 200 Nm-1

kE = 10 / 0.033 = 303 Nm-1

The largest constant is 320 Nm-1.

11. The graph below show connection between the change in force (ΔF) and the increase in length (Δx). What is the graph showing the smallest constant of elasticity.

Solution

The equation of Hooke’s law :

k = F / Δx

Δx = the change in length, F = force, k = constant of elasticity

Constant of elasticity :

kA = F / Δx = 1 / 8 = 0.125

kB = F / Δx = 8 / 3 = 2.7

kC = F / Δx = 6 / 6 = 1

kD = F / Δx = 3 / 5 = 0.6

kE = F / Δx = 2 / 4 = 0.5

12. Which graph has the largest elastic constants?

Solution :

Constant of elasticity :

kA = F / Δx = 50 / 10 = 5

kB = F / Δx = 50 / 0.1 = 500

kC = F / Δx = 5 / 0.1 = 50

kD = F / Δx = 500 / 0.1 = 5000

kE = F / Δx = 500 / 10 = 50

The potential energy of spring :

13.The graph below shows the relationship between force and the change in spring length. What is the potential energy of spring, according to the graph below.

Known :

F = 40 N

x = 0.08 meters

Wanted : The potential energy of spring

Solution :

Constant of spring :

k = F / Δx = 40 / 0.08 = 500 N/m

The potential energy of spring :

PE = 1/2 k x2 = 1/2 (500)(0.08) = (250)(0.08) = 20 Joule

14. A 2-kg block is attached at spring. If the increase in length of spring is 5 cm and acceleration due to gravity is 10 m/s2, what is the potential energy of spring.

Known :

The increase in length (Δx) = 5 cm = 0.05 meter

Acceleration due to gravity (g) = 10 m/s2

Block’s mass (m) = 2 kg

Block’s weight (w) = m g = (2)(10) = 20 Newton

Wanted : the potential energy of spring

Solution :

The constant of elasticity :

k = w / Δx = 20 / 0.05 = 400 N/m

The potential energy of spring :

PE = ½ k Δx2 = ½ (400)(0.05)2 = (200)(0.0025)

PE = 0.5 Joule

15. The change in length of spring is 5 cm when pulled by 20-N force. What is the potential energy of spring when the change in length of the spring is 10 cm.

Known :

The change in length (Δx) = 5 cm = 0.05 meters

Force (F) = 20 Newton

Wanted : The potential energy of spring

Solution :

The constant of spring :

k = F / Δx = 20 / 0.05 = 400 N/m

The potential energy of spring when Δx = 10 cm = 0.1 m :

PE = ½ k Δx2 = ½ (400)(0.1)2 = (200)(0.01)

PE = 2 Joule

Object’s weight

16. Four springs where the constant of each spring is 800 N/m, connected in series-parallel, as shown in figure. A block is attached at spring. The change in length of all springs is 5 cm. What is the weight of block.

Known :

k1 = k2 = k3 = k4 = 800 N m−1

Δx = 5 cm = 0.05 m

Wanted : block’s weight (w)

Solution :

Determine the constant of the equivalent spring

Spring 1 (k1), spring 2 (k2) and spring 3 (k3) are connected in parallel. The constant of the equivalent spring :

kp = k1 + k2 + k3 = 800 + 800 + 800 = 2400 Nm−1

Spring p (kP) and spring 4 (k4) are connected in series. The constant of the equivalent spring :

1/ks = 1/kp + 1/k4 = 1/2400 + 1/800 = 1/2400 + 3/2400 = 4/2400

ks = 2400/4 = 600 Nm−1

The constant of the equivalent spring is 600 Nm-1

Determine the weight of object :

The equation of Hooke’s law :

F = k Δx tau w = k Δx

Weight of object :

w = (600 Nm-1)(0.05 m) = 30 Newton

17. Four springs are connected in series-parallel. Constant of each spring is 1600 N/m. A block is attached at the end of spring, as shown in figure. The increase in length of all spring is 5 cm. What is the weight of block.

Known :

k1 = k2 = k3 = k4 = 1600 N m−1

Δx = 5 cm = 0.05 m

Wanted : weight of block

Solution :

Determine the constant of the equivalent spring

Spring 1 (k1), spring 2 (k2) and spring 3 (k3) are connected in parallel. The constant of the equivalent spring :

kP = k1 + k2 + k3 = 1600 + 1600 + 1600 = 4800 Nm−1

Spring p (kP) and spring 4 (k4) are connected in series. The constant of the equivalent spring :

1/ks = 1/kp + 1/k4 = 1/4800 + 1/1600 = 1/4800 + 3/4800 = 4/4800

ks = 4800/4 = 1200 Nm−1

The constant of the equivalent spring is 1200 Nm-1

Determine the weight of object :

The equation of Hooke’s law :

F = k Δx or w = k Δx

Object’s weight :

w = (1200 Nm-1)(0.05 m) = 60 Newton

1. What is Hooke’s Law?
• Answer: Hooke’s Law describes the relationship between the force applied to an elastic object and the resulting deformation (usually elongation or compression). Specifically, it states that the force required to compress or extend a spring is directly proportional to the distance it is stretched or compressed, provided the elastic limit is not exceeded.
2. What does it mean when we say a material has reached its elastic limit?
• Answer: When a material has reached its elastic limit, it means that it will no longer return to its original shape or size after the deforming force is removed. Beyond this point, the material behaves plastically and may be permanently deformed.
3. How does the spring constant (k) relate to the stiffness of a spring?
• Answer: The spring constant (k) is a measure of a spring’s stiffness. A larger value of k indicates a stiffer spring, meaning more force is needed to deform it a given amount, while a smaller k indicates a more compliant or softer spring.
4. What are the units of the spring constant in the SI system?
• Answer: In the SI system, the units for the spring constant (k) are Newtons per meter (N/m).
5. Why is the behavior described by Hooke’s Law considered linear?
• Answer: The behavior is considered linear because the relationship between the force applied (F) and the displacement (x) is a straight line, with the relationship given as , where k is a constant for a given material or spring.
6. Is Hooke’s Law applicable only to springs?
• Answer: No, Hooke’s Law is applicable to any elastic material that deforms linearly with the applied force, up to its elastic limit. While springs are a common example, other materials such as rubber bands, metals under small deformations, and some biological tissues can also exhibit behavior described by Hooke’s Law.
7. What happens if a material is stretched beyond its elastic limit but not enough to break?
• Answer: If a material is stretched beyond its elastic limit but not to the point of breaking, it will undergo plastic deformation. This means that when the force is removed, the material will not return entirely to its original shape, and some permanent deformation will remain.
8. How do the concepts of stress and strain relate to Hooke’s Law?
• Answer: Stress is the force applied per unit area, and strain is the relative deformation of a material. Hooke’s Law in terms of stress and strain states that the stress is directly proportional to the strain, with the proportionality constant being the material’s Young’s modulus. This is another way of expressing the linear relationship between force and deformation, but for bulk materials rather than just springs.
9. What is Young’s Modulus and how does it relate to elasticity?
• Answer: Young’s Modulus, represented usually by the letter , is a measure of a material’s stiffness in terms of tension or compression. It describes the material’s ability to resist deformation under an applied force. A higher Young’s Modulus indicates a stiffer material, and it’s defined as the ratio of stress to strain.
10. Can all materials be described by Hooke’s Law?
• Answer: No, not all materials behave according to Hooke’s Law. Many materials, especially those that are non-linear, viscoelastic, or plastic, do not exhibit a linear relationship between stress and strain. Hooke’s Law is an idealized description and is most accurate for small deformations of elastic materials.