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A spring with spring constant \(\mathrm{K}\) when stretched through \(2 \mathrm{~cm}\) the potential energy is \(\mathrm{U}\). If it is stretched by \(6 \mathrm{~cm}\). The potential energy will be...... (A) \(6 \mathrm{U}\) (B) \(3 \mathrm{U}\) (C) \(9 \mathrm{U}\) (D) \(18 \mathrm{U}\)

Short Answer

Expert verified
The potential energy of the spring when stretched by 6 cm is \( 9 \mathrm{U} \).

Step by step solution

01

Write the formula for potential energy of a spring

The potential energy U of a spring is given by the formula: \[ U = \frac{1}{2} kx^2 \] Here, U = potential energy k = spring constant x = extension of the spring
02

Find the relationship of the potential energy when the extension is doubled

When the spring is stretched by 2 cm, the potential energy is U. We can represent this situation as: \[ U = \frac{1}{2} k(2)^2 = 2k \] Now, the problem asks us what happens when the spring is stretched by 6 cm. We can represent this situation as: \[ U' = \frac{1}{2} k(6)^2 = 18k \]
03

Calculate the ratio of the new potential energy to the existing potential energy

To obtain the relationship between the new potential energy U' and the given potential energy U, we can divide the expression for U' by the expression for U: \[ \frac{U'}{U} = \frac{18k}{2k} = \frac{18}{2} = 9 \] Thus, the new potential energy U' is 9 times the initial potential energy U: \[ U' = 9U \] So, the correct answer is (C) \(9 \mathrm{U}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Constant
In the world of springs, the spring constant, denoted by \( k \), is a crucial value that defines the stiffness of the spring. The stiffer the spring, the larger its spring constant. You can think of the spring constant as the resistance a spring offers when you try to compress or stretch it. This value helps us understand how much force is needed to extend or compress the spring by a certain amount.

Mathematically, if you apply a force \( F \) to the spring and it stretches by a distance \( x \), the spring constant \( k \) can be calculated using the formula:
  • \( F = kx \)
This equation tells us that force is directly proportional to the distance the spring is stretched or compressed. This relationship is a key factor in finding out the potential energy stored in a spring.
Hooke's Law
Hooke's Law is a fundamental principle in understanding how springs behave. It states that the force required to compress or extend a spring by a distance \( x \) from its natural length is directly proportional to that distance. In simpler terms, the further you stretch a spring, the more force it takes to stretch it even further. This law is expressed through the formula:
  • \( F = kx \)
Where:
  • \( F \) stands for the force applied
  • \( k \) is the spring constant
  • \( x \) is the extension or compression of the spring
The proportionality persists only within the elastic limit of the material. Beyond this limit, the material might deform permanently and Hooke's law doesn't hold. The principle helps us predict how a spring will behave under certain forces, making it easier to calculate the potential energy stored in the spring.
Energy in Elastic Potential
Energy in elastic potential is the energy stored in objects when they are stretched or compressed, like springs. This stored energy can be released as kinetic energy when the object returns to its original shape. In a spring, the potential energy is given by the formula:
  • \( U = \frac{1}{2} kx^2 \)
Where:
  • \( U \) is the elastic potential energy
  • \( k \) is the spring constant
  • \( x \) is the amount of stretch or compression from the spring's equilibrium position
The energy equation tells us that the potential energy increases with the square of the displacement \( x \). Therefore, tripling the extension of a spring will increase the potential energy by nine times \((3^2 = 9)\). This concept was used in the original exercise to determine the spring's potential energy when stretched to a different length. Understanding how energy is stored and released helps in various applications such as shock absorbers, watches, and even trampolines.

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Most popular questions from this chapter

Four identical balls are lined in a straight grove made on a horizontal frictionless surface as shown. Two similar balls each moving with a velocity v collide elastically with the row of 4 balls from left. What will happen (A) One ball from the right rolls out with a speed \(2 \mathrm{v}\) and the remaining balls will remain at rest. (B) Two balls from the right roll out speed \(\mathrm{v}\) each and the remaining balls will remain stationary. (C) All the four balls in the row will roll out with speed \(\mathrm{v}(\mathrm{v} / 4)\) each and the two colliding balls will come to rest. (D) The colliding balls will come to rest and no ball rolls out from right.

A body is moved along a straight line by a machine delivering a constant power. The velocity gained by the body in time \(t\) is proportional to..... (A) \(t^{(3 / 4)}\) (B) \(t^{(3 / 2)}\) (C) \(t^{(1 / 4)}\) (D) \(t^{(1 / 2)}\)

A mass of \(\mathrm{M} \mathrm{kg}\) is suspended by a weight-less string, the horizontal force that is required to displace it until the string makes an angle of \(60^{\circ}\) with the initial vertical direction is (A) \(\mathrm{Mg} / \sqrt{3}\) (B) \(\mathrm{Mg} \cdot \sqrt{2}\) (C) \(\mathrm{Mg} / \sqrt{2}\) (D) \(\mathrm{Mg} \cdot \sqrt{3}\)

A force of \((2 \mathrm{i} \wedge+3 j \wedge-\mathrm{k} \wedge) \mathrm{N}\) acts on a body for 5 second, produces a displacement of \((3 i \wedge+5 j \wedge+k \wedge)\). What was the power used? (A) \(4 \mathrm{~W}\) (B) \(20 \mathrm{~W}\) (C) \(21 \mathrm{~W}\) (D) \(4.2 \mathrm{~W}\)

A gun fires a bullet of mass \(40 \mathrm{~g}\) with a velocity of \(50 \mathrm{~m} / \mathrm{s}\). Because of this the gun is pushed back with a velocity of \(1 \mathrm{~m} / \mathrm{s}\). The mass of the gun is (A) \(1.5 \mathrm{~kg}\) (B) \(3 \mathrm{~kg}\) (C) \(2 \mathrm{~kg}\) (D) \(2.5 \mathrm{~kg}\)

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