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

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2. Hooke's law and elasticity – problems and solutions 1There 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 :

Hooke's law and elasticity – problems and solutions 2Determine 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

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The constant of spring

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

Hooke's law and elasticity – problems and solutions 3

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 :Hooke's law and elasticity – problems and solutions 4

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.

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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 :

Hooke's law and elasticity – problems and solutions 5

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 :

Hooke's law and elasticity – problems and solutions 6

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

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

Solution :Hooke's law and elasticity – problems and solutions 7

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 :

Hooke's law and elasticity – problems and solutions 8

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…

Hooke's law and elasticity – problems and solutions 9

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

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

Hooke's law and elasticity – problems and solutions 10

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.

Hooke's law and elasticity – problems and solutions 11

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?

Hooke's law and elasticity – problems and solutions 12

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

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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 :Hooke's law and elasticity – problems and solutions 13

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 :Hooke's law and elasticity – problems and solutions 14

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

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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 :Hooke's law and elasticity – problems and solutions 15

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

Hooke's law and elasticity – problems and solutions 16Spring 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 :Hooke's law and elasticity – problems and solutions 17

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

Hooke's law and elasticity – problems and solutions 18Spring 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