1. A 2-kg mass is attached to a spring. If the elongation of spring is 4 cm, determine potential energy of elastic spring. Acceleration due to gravity is 10 m/s2.
Known :
Mass (m) = 2 kg
Acceleration due to gravity (g) = 10 m/s2
Weight (w) = m g = (2)(10) = 20 N
Elongation (x) = 4 cm = 0.04 m
Wanted : Potential energy of elastic spring
Solution :
Formula of Hooke’s law :
F = k x
F = force, k = spring constant, x = the change in length of spring
Spring constant :
k = w / x = 20 / 0.04 = 500 N/m
Potential energy of elastic spring :
PE = ½ k x2 = ½ (500)(0.04)2 = (250)(0.0016) = 0.4 Joule
Alternative solution :
PE = ½ k x2 = ½ (w / x) x2 = ½ w x = ½ m g x
w = weight, m = mass, x = the change in length of spring
PE = ½ (2)(10)(0.04) = (10)(0.04) = 0.4 Joule.
2. The elongation of spring stretched by a force F = 50 N is 10 cm. What is the potential energy of elastic spring if the elongation of spring is 12 cm.
Known :
Elongation (x) = 10 cm = 0.1 m
Force (F) = 50 N
Wanted : The potential energy of elastic spring
Solution :
Spring constant :
k = F / x = 50 / 0.1 = 500 N/m
The potential energy of elastic spring if the elongation of spring is 12 cm. :
PE = ½ k x2 = ½ (500)(0.12)2 = (250)(0.0144) = 3.6 Joule.
3. Graph of F vs x shown in figure below. What is the potential energy of elastic spring, if the elongation of spring is 10 cm.
Known :
Force (F) = 5 N
The elongation of spring (x) = 2 cm = 0.02 m
Wanted: The potential energy of elastic spring if the elongation of spring is 0.1 m.
Solution :
Spring constant :
k = F / x = 5 / 0.02 = 250 N/m
The potential energy of elastic spring if the elongation of spring is 0.1 m :
PE = ½ k x2 = ½ (250)(0.1)2 = (125)(0.01) = 1.25 Joule.
4. An athlete jumps onto a spring instrument with a weight of 500 N, the spring shortens 4 cm. Determine the amount of the potential energy to force the athlete.
A. 20 Joule
B. 10 Joule
C. 5 Joule
D. 2 Joule
Known :
Force (F) = 500 N
The change in length of spring (Δx) = 4 cm = 0.04 m
Wanted: The potential energy of spring
Solution :
Calculate the elastic constant of a spring using the equation of Hooke’s law :
k = F / Δx = 500 N / 0.04 m = 12500 N/m
The elastic potential energy of spring :
PE = ½ k Δx2 = ½ (12500)(0.04)2 = (6250)(0.0016)
PE = 10 Joule
The correct answer is B.
5. A spring when suspended with a mass of 500 grams, its length increases by 5 cm. Determine its constant and its potential energy.
Known :
Mass of object (m) = 500 gram = (500/1000) kg = 5/10 kg = 0.5 kg
Acceleration (g) = 10 m/s2
Object weight (w) = m g = (0.5 kg)(10 m/s2) = 5 kg m/s2 = 5 Newton
The change in length of spring (Δx) = 5 cm = 0.05 m
Wanted : spring’s constant (k) and potential energy (PE)
Solution :
First, calculate spring’s constant using equation of Hooke’s law :
k = w / Δx = 5 N / 0.05 m = 100 N/m
The potential elastic of spring :
PE = ½ k Δx2 = ½ (100)(0.05)2 = (50)(0.0025)
PE = 0.125 Joule
[wpdm_package id=’1179′]
- Work done by force problems and solutions
- Work-kinetic energy problems and solutions
- Work-mechanical energy principle problems and solutions
- Gravitational potential energy problems and solutions
- Potential energy of elastic spring problems and solutions
- Power problems and solutions
- Application of conservation of mechanical energy for free fall motion
- Application of conservation of mechanical energy for up and down motion in free fall motion
- Application of conservation of mechanical energy for motion on curve surface
- Application of conservation of mechanical energy for motion on inclined plane
- Application of conservation of mechanical energy for projectile motion
