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
[irp]
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
[irp]
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.
[irp]
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
[irp]
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
[irp]
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
[irp]
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
[irp]
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
