Last updated on Mar 20, 2025
When a spring is stretched through a distance s, its potential energy is 10 Joules. The work (in Joule) required to stretch it further through s will be :
CONCEPT:
\(\Rightarrow W = \frac{1}{2} k x^{2}\)
Where K = Force constant and x = Displacement of the spring
CALCULATION :
Given U = 10 Joules, x1 = x, and x2 = 2 x
\(\Rightarrow 10J = \frac{1}{2} k x^{2}\)
\(\Rightarrow W_{2} = \frac{1}{2}k (2x)^{2}\)
\(\Rightarrow W_{2} = 2kx^{2 }\)
\(\Rightarrow W =W_{2} -10 J\)
Substituting the given values in the above equation
\(\Rightarrow W = 2kx^{2} - \frac{1}{2}kx^{2} = \frac{3}{2}kx^{2}\)
\(\Rightarrow W = 3 \times \frac{1}{2}k x^{2}\)
\(\Rightarrow W = 3 \times 10 = 30 J\)
A spring block system stretched and then left to move. At which point the block will have maximum speed?
CONCEPT:
The Kinetic Energy of an object due to its linear speed is given by:
E = (1/2)(m × v2)
where m is the mass of a body and v is the speed.
The potential energy of a spring is given by:
\(PE = {1\over 2}kx^2\)
where k is the spring constant, x is the elongation or compression in the spring.
Total initial mechanical energy = Total final mechanical energy
EXPLANATION:
\(PE = {1\over 2}kx^2\)
\(0 = {1\over 2}kx^2\) i.e. x = 0
Two springs with a spring constant of 400 N/m and 100 N/m are stretched in such a way that both have same potential energy. What will be the ratio of elongation in first spring to the second spring?
CONCEPT:
When a spring is stretched by length 'x', potential energy stored in it is given by:
\(U = {1\over 2}kx^2\)
Where k is a spring constant and x is the displacement from the mean position.
CALCULATION:
Given that both springs have the same potential energy and k1 = 400 N/m and k2 = 100 N/m
\(U_1 = U_2\)
\( {1\over 2}k_1x_1^2= {1\over 2}k_2x_2^2\)
\( {x_1^2\over x_2^2}={100 \over 400}\)
\( {x_1\over x_2}={1\over 2}\)
So the correct answer is option 2.
In an ideal spring, the spring force F, is proportional to x where x is the displacement of the block from the equilibrium position. The displacement could be either positive or negative. This force law for the spring is called Hooke's law and is mathematically stated as:
Concept:
Spring force: In an ideal spring, the force required to stretch a string from its equilibrium position is directly proportional to the extension of the spring.
This is known as Hooke's law and commonly written:
Fs = -kx
Where Fs = Spring force, x = Displacement from the equilibrium position and k is the spring constant.
Explanation:
The Hooke's law for the spring is given by:
Fs = -kx
The slider of mass m = 5 kg is released from rest in position A and slides down the vertical plane guide. A constant downward F = 5 N acts on the slider vertically throughout the motion. If the maximum compression of the spring is to be 40 mm, determine the work done by friction. Spring constant, K = 18,000 N/m
Concept:
The potential energy of the spring = \( {{1} \over 2}× k\times (Δ x)^2\)
Work done by gravity force = F× distance along the force = m × a × distance along the force
Work done by the constant force = F × distance along the force
Calculation:
Given:
Mass of the slider = 5 kg
Constant downward force F= 5 N
Compression of the spring (Δx) = 40 mm = 0.04 m
Work done by gravity force = m × a × distance along the force = 5 × 10 × 0.6 = 30 J
Work done by the constant force = force × distance along the force = 5 × 0.6 = 3 J
Work done by friction force = x
Then total work done = 30 + 3 + x = 33 + x
Energy stored in the spring = \( {{1} \over 2}× 18000\times (0.04)^2= 14.4~J\)
Total work done = potential energy of the spring
33 + x = 14.4
x = -18.6 J
Which of the following gives correct equation for Hooke's law for the spring? (F is force, k is spring constant and x is extension/compression in spring)
CONCEPT:
This is known as Hooke's law and commonly written:
Fs = -kx
Where Fs is the spring force, x is the displacement from the equilibrium position and k is the spring constant.
EXPLANATION:
The Hooke's law for the spring is given by:
Fs = -kx
What will be the spring constant of a spring? When it is streched 10 cm, it has potential energy of 5600 J.
Calculation:
The potential energy stored in a stretched spring is given by the formula:
\(U = \frac{1}{2} k x^2 \)
where;
U is the potential energy (5600 J),
k is the spring constant (to be determined),
x is the extension or compression of the spring in meters (10 cm = 0.1 m).
then;
\(k = \frac{2U}{x^2}\)
\(k = \frac{11200}{0.01} = 1.12 \times 10^6 \, \text{N/m}\)
The spring constant is: \(k = 1.12 \times 10^6 \, \text{N/m}\).
Thus, option '3' is correct.
A spring, which is initially in its unstretched condition, is first stretched by a length x and then again by a further length x. The work done in the first case is W1 and in the second case is W2.
Concept:
The work done by the force is defined as \(\text{W}=\int F \cdot dx\) , where F is the force and \(\text{dx}\) is the line element.
Explanation:
A spring, which is initially in its unstretched condition, It is first stretched by x m and its work done is given as
\(\text{W}_{1}=\int_0^xkxdx \\ \text{W}_{1}=\frac{1}{2}kx^2\)
Then work done in moving further from x to 2x position such that the next stretched length is also x m,
\(\text{W}_{2}=\int_x^{2x}kxdx \\ \text{W}_{2}=\frac{k}{2}[4x^2-x^2] \\ \text{W}_{2}=\frac{3}{2}kx^2\)
Then the ratio of \(\frac{\text{W}_{2}}{\text{W}_{1}}=3\).
The correct option is (3) \(\text{W}_{2}=3\text{W}_{1}\) .
The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be:
Concept:
Potential Energy in a Spring:
Calculation:
Given,
Initial displacement, \(x_1 = 2\) cm = 0.02 m
Potential energy at \(x_1\), \(U_1 = U\)
New displacement, \(x_2 = 8\) cm = 0.08 m
Potential energy at \(x_2\), \(U_2\)
The potential energy stored in the spring is given by:
\(U = \frac{1}{2} k x^2\)
For the initial displacement:
\(U_1 = \frac{1}{2} k x_1^2\)
For the new displacement:
\(U_2 = \frac{1}{2} k x_2^2\)
Using the given displacements:
\(U_2 = \frac{1}{2} k (0.08)^2\)
\( U_2 = \frac{1}{2} k (4 \times 0.02)^2\)
\( U_2 = \frac{1}{2} k (16 \times 0.02^2)\)
\( U_2 = 16 \times \frac{1}{2} k (0.02)^2\)
\( U_2 = 16 \times U_1\)
∴ The potential energy stored in the spring when stretched by 8 cm is 16U.
Work done by spring force depends on _______.
CONCEPT:
This is known as Hooke's law for springs:
Fs = -kx
Where Fs is the spring force, x is the displacement from the equilibrium position and k is the spring constant.
Work done by a variable force is given by:
\(W=\int_{x_1}^{x_2}F . dx\)
where W is the work done. F is the force, dx is the displacement, x1 and x2 are the limits from which the body moves to which the body moves.
EXPLANATION:
We know that spring force is given by:
Fs = -kx
So work done by it for moving an object from x1 to x2
\(W=\int_{x_1}^{x_2}F . dx=\int_{x_1}^{x_2}(-kx) . dx=[{-kx^2 \over 2}]_{x_1}^{x_2}\)
\(W={kx_1^2 \over 2}-{kx_2^2 \over 2}\)