# 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 (F_{1}) = F

The change in length 1 (∆L_{1}) = ∆L

The change in length 2 (∆L_{2}) = 4 ∆L

__Wanted :__ Force 2 (F_{2})

__Solution :__

The equation of Hooke’s law

k = F / ΔL

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

k_{1 }= k_{2}

F_{1} / ∆L_{1} = F_{2 }/ ∆L_{2}

F / ΔL = F_{2 }/ 4ΔL

F / 1 = F_{2 }/ 4

F = F_{2 }/ 4

F_{2 }= 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/s^{2}. What is the change in length of the equivalent spring.

__Known :__

Object’s mass (m) = 100 gram = 0.1 kg

k_{1} = k_{2} = k_{3} = 200 N/m

w = m g = (0.1 kg)(10 m/s^{2}) = 1 kg m/s^{2} = 1 Newton

__Wanted :__ The change in length of the equivalent spring.

__Solution :__

Determine the equivalent spring constant :

Spring 2 (k_{2}) and spring 3 (k_{3}) are connected in parallel. The equivalent spring constant :

k_{p} = k_{2} + k_{3} = 200 + 200 = 400 Nm^{−1}

Spring 1 (k_{1}) and spring p (k_{P}) are connected in series. The equivalent spring constant :

1/k_{s} = 1/k_{p} + 1/k_{1} = 1/400 + 1/200 = 1/400 + 2/400 = 3/400

k_{s} = 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 k_{1} = k_{2 }= 3 Nm^{−1} and k_{3} = 6 Nm^{−1}. What is the constant of the equivalent of spring.

__Known :__

Constant of spring 1 (k_{1}) = constant of spring 2 (k_{2}) = 3 Nm^{−1 }

Constant of spring 3 (k_{3}) = 6 Nm^{−1 }

__Wanted :__ constant of the equivalent spring (k)

__Solution :__

Spring 1 (k_{1}) and spring 2 (k_{2}) are connected in parallel. The Constant of the equivalent spring :

k_{p} = k_{1} + k_{2} = 3 + 3 = 6 Nm^{−1}

Spring p (k_{P}) and spring 3 (k_{3 }) are connected in series. The constant of the equivalent spring :

1/k_{s} = 1/k_{p} + 1/k_{ 3} = 1/6 + 1/3 = 1/6 + 2/6 = 3/6

k_{s} = 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 k_{1} = 4k, what is the constant of the equivalent spring.

Solution :

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

k_{p} = k + k = 2k

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

1/k_{s} = 1/k_{p} + 1/k_{1} = 1 / 2k + 1 / 4k = 2 / 4k + 1 / 4k = 3 / 4k

k_{s} = 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 / *Δ**L *

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 :

k_{A} = F / Δx = 1 / 0.05 = 20 N/m

k_{B} = F / Δx = 2 / 0.025 = 80 N/m

k_{C} = F / Δx = 1 / 0.025 = 40 N/m

k_{D} = F / Δx = 2 / 0.05 = 40 N/m

k_{E} = 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

k_{A} = 7 / 0.035 = 200 Nm^{-1}

k_{B} = 8 / 0.025 = 320 Nm^{-1}

k_{C} = 6 / 0.020 = 300 Nm^{-1}

k_{D} = 9 / 0.045 = 200 Nm^{-1}

k_{E} = 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 elasticit

Solution

__The equation of Hooke’s law :__

k = F / Δx

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

__Constant of elasticity :__

k_{A} = F / Δx = 1 / 8 = 0.125

k_{B} = F / Δx = 8 / 3 = 2.7

k_{C} = F / Δx = 6 / 6 = 1

k_{D} = F / Δx = 3 / 5 = 0.6

k_{E} = F / Δx = 2 / 4 = 0.5

12. Which graph has the largest elastic constants?

Solution :

__Constant of elasticity :__

k_{A} = F / Δx = 50 / 10 = 5

k_{B} = F / Δx = 50 / 0.1 = 500

k_{C} = F / Δx = 5 / 0.1 = 50

k_{D} = F / Δx = 500 / 0.1 = 5000

k_{E} = 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 x^{2 }= 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/s^{2}, 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/s^{2}

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 Δx^{2 }= 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 Δx^{2 }= 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 :__

k_{1} = k_{2} = k_{3} = k_{4 }= 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 (k_{1}), spring 2 (k_{2}) and spring 3 (k_{3}) are connected in parallel. The constant of the equivalent spring :

k_{p} = k_{1} + k_{2} + k_{3} = 800 + 800 + 800 = 2400 Nm^{−1}

Spring p (k_{P}) and spring 4 (k_{4}) are connected in series. The constant of the equivalent spring :

1/k_{s} = 1/k_{p} + 1/k_{4} = 1/2400 + 1/800 = 1/2400 + 3/2400 = 4/2400

k_{s} = 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 :__

k_{1} = k_{2} = k_{3} = k_{4 }= 1600 N m^{−1}

Δx = 5 cm = 0.05 m

__Wanted :__ weight of block

__Solution :__

__Determine the constant of the equivalent spring__

Spring 1 (k_{1}), spring 2 (k_{2}) and spring 3 (k_{3}) are connected in parallel. The constant of the equivalent spring :

k_{P} = k_{1} + k_{2} + k_{3} = 1600 + 1600 + 1600 = 4800 Nm^{−1}

Spring p (k_{P}) and spring 4 (k_{4}) are connected in series. The constant of the equivalent spring :

1/k_{s} = 1/k_{p} + 1/k_{4} = 1/4800 + 1/1600 = 1/4800 + 3/4800 = 4/4800

k_{s} = 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