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When an object repeats its motion in a predetermined cycle, it is said to be in periodic motion. Oscillation is another term for this sort of motion. Spring and pendulum movement are simple examples, but there are numerous additional circumstances in which oscillations occur. One of the most significant characteristics of periodic motion is that the item maintains a steady equilibrium position. A restoring force is also at work for string, which means that string has potential energy.
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Key Takeaways: Motion, Elastic Potential Energy, Deformation, Compression, Equilibrium Point, Force, Constant, Displacement
Spring Potential Energy
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Spring is a popular tool, and its inertia is commonly overlooked because of its negligible mass. When spring is stressed, it deforms because of compression. Then it reaches its equilibrium point. As a result, a spring exerts an equal and opposite force when it compresses or expands a body.
A compressible or stretchable object, such as a spring, rubber band, or molecule, stores energy. It is also known as elastic potential energy. It's the force multiplied by the movement's distance.
The spring has no energy when it is in its usual position, which is when it is not stressed. However, if we shift the spring from its usual place, the spring will be able to retain energy because of its new location. Potential energy is the name given to this type of energy as it is caused by the deformation of a certain elastic item, such as a spring. This energy also refers to the process of stretching the spring. It is determined by the spring constant 'k' and the stretched distance.
Spring Potential Energy Formula
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String potential energy is calculated by the product of force and distance of displacement, where the force is equal to the product of spring constant and spring displacement.
\(P.E. = \frac{1}{2} k \times x^2\)
Here,
P.E = Spring Potential Energy
k= Spring Constant
χ= Spring Displacement
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Points to Remember
Following are the important points:
- The spring's potential energy is the energy stored because of the deformation of a certain elastic item, such as a spring. It depends on the spring constant k and the distance stretched to define the labour done to stretch the spring.
- The spring constant is the spring's distinguishing feature. It is measured in units of N m-1 and is dependent on the construction material.
- The spring force is an example of a conservative variable force. The spring force Fs is proportional to x in an ideal spring, where x is the displacement of the block from its equilibrium position.
- The displacement may be positive or negative. Hooke's law is the force law that governs the spring.
- In a cyclic process, the work done by the spring force is zero. As a result, the spring force is conservative.
Sample Questions
Ques: A stretched string has a spring constant of 50Nm-1 and a displacement of 20 cm. Calculate the stored potential energy in the stretched string. (2 Marks)
Ans: We know that,
Potential spring energy or \(P.E. = \frac{1}{2} k \times x^2\)
=\(P.E. = \frac{1}{2} 50 \times (0.2)^2\)
= 1 J
As a result, the stored potential energy in the stretched string will be 1 joule.
Ques: A spring is linked to a 2-kg mass. Determine the elastic spring's potential energy if its elongation is 4 cm. Gravitational acceleration is 10 m/s2. (2 Marks)
Ans: \(P.E. = \frac{1}{2} k \times x^2\) =\(P.E. = \frac{1}{2}(\frac{w}{x}) x^2\)
=\(\frac{1}{2}w x\) =\(\frac{1}{2}mgx\)
PE = (2)(10)(0.04) = (10)(0.04) = 0.4 Joule.
As a result, the stored potential energy in the stretched string will be 0.4 joule.
Ques: Calculate the potential energy of spring with a 200 N/m spring constant and a displacement of 0.8 m. (2 Marks)
Ans: Potential energy of a string formula is given by,
P.E = 1/2 kx2
P.E = ½ (200 x (0.8)2)
P.E = 64 J
As a result, the stored potential energy in the stretched string will be 64 joules.
Ques: A stretched string's spring constant and displacement are 100 N/m and 0.5 m, respectively. Calculate the amount of potential energy in the stretched string. (2 Marks)
Ans: Potential energy of a string formula is given by,
P.E = 1/2 kx2
P.E = 1/2 x (100 x (0.5)2)
P.E = 12.5 J
As a result, the stored potential energy in the stretched string will be 12.5 joule.
Ques: A graph of F vs. x is shown in the diagram below. What is the potential energy of a 10 cm elongated elastic spring? (3 Marks)
Ans: Spring constant :
k = F / x (as per Hooke’s Law)
= 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.
As a result, the stored potential energy in the stretched string will be 1.25 joule.
Ques: We attach a spring to a board and stretch it 99 cm using 3 J of energy. Using the Spring Potential Energy Formula, what is the value of the spring constant? (3 Marks)
Ans: Work done = P.E. = 3 J
Displacement, x = 99 cm = 0.99 m.
\(P.E. = \frac{1}{2}(\frac{w}{x}) x^2\)
Rearranging it,
\(k = \frac{2 \times P.E.}{x^2}\)
\(k = \frac{2 \times 3.}{0.99^2}\)
=6.122Nm-1
Thus, the spring constant is 6.122Nm-1.
Ques: Determine the ball's velocity shortly before it hits the ground. Assume the ball started at a height of 100 metres and a mass of 4 kg. (4 Marks)
Ans: At first, the ball has potential energy at a height of 10 metres. When it is dropped, it begins to fall to the earth and its height begins to decrease. The velocity of the object rises as the height decreases, and it gains kinetic energy.
Potential Energy at t is 0
Potential energy, P = mgh
m = 4Kg
h = 100m
g = 10 m/s2
P = (4)(100)(10)
P = 4000J
When the ball is going to touch the ground, its potential energy has vanished, and all the energy has been transformed to kinetic energy.
Thus, K.E = 4000J
\(K.E. = \frac{1}{2} mv^2\)
M= 4 kgs and K.E. = 4000J
4000J = \(\frac{1}{2} 4v^2\)
2000= v2
v= 20\(\sqrt{5}\)m/s
Ques: Determine the ball's kinetic energy shortly before it hits the ground. Assume the ball started at a height of 10 metres and a mass of 2 kg. (3 Marks)
Ans: At first, the ball has potential energy at a height of 10 metres. When it is dropped, it begins to fall to the earth and its height begins to decrease. The velocity of the object rises as the height decreases, and it gains kinetic energy.
Potential Energy at t is 0
Potential energy = mgh
m = 2Kg, h = 10m and g = 10 m/s2
P = (2)(10)(10)
P = 200J
When the ball is going to touch the ground, its potential energy has vanished, and all the energy has been transformed into kinetic energy.
Hence, K.E= 200J
Ques: The length of the spring rises by 5 cm when hung with a load of 500 grams. Determine the constant and potential energy of the object. (3 Marks)
Ans: Calculating 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 :
\(P.E. = \frac{1}{2}k\bigtriangleup x^2\)
=12(100)(0.05)2
= (50)(0.0025)
PE = 0.125 Joule
Ques: If the spring's potential energy is V after extending it by 2 cm, what is its potential energy after stretching it by 10 cm? (3 Marks)
Ans: \(P.E. = \frac{1}{2}k\times x^2\)
In the formula mentioned above, k denotes the spring's spring constant in N/m, and x is the spring's displacement from its equilibrium position in metres. As per the question, the potential energy is V when the displacement is 2cm (0.02m).
\(V = \frac{1}{2}k\times 0.02^2\)
As displacement is 10 cm (0.1 m)
\(U = \frac{1}{2}k\times 0.1^2\)
Dividing the above two equations
\(\frac{U}{V} = \frac{0.01}{0.0002}\)
U= 25V
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