Answer
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Hint:The conservation of energy is the most fundamental law in the universe. When the mass falls from a certain height, it carries its potential energy with it. The spring compresses by a certain length and the mass comes to rest at that point. The gravitational potential energy is converted to elastic potential energy in the spring. So,
\[mg(h + x) = \dfrac{1}{2}k{x^2}\]
Complete step by step answer:
Let the compression in the spring be x. The point where the upper end of spring comes along with the mass of 2 kg is the level of zero gravitational potential.
Two of the most fundamental laws applicable in the whole universe are the Conservation of Energy and Conservation of Momentum. Here the conservation of Energy is applicable. The mass stored with some gravitational energy comes to the point of zero potential which is defined above. The gravitational potential is converted and stored as elastic potential energy in the spring. The spring is compressed by a length x.
The elastic potential energy of a spring is given as:
\[\Rightarrow {U_1} = \dfrac{1}{2}k{x^2}\]
The mass moves a distance of 40 cm along with a distance x. This is the same distance the spring compresses as the mass comes down along with the top end of the spring. So, the gravitational potential energy stored in the mass is:
\[\Rightarrow {U_2} = mg(h + x)\]
Equating the two equations and substituting the data from the question (let g=10m/s.s for simplicity), we have,
\[
\Rightarrow \dfrac{1}{2}1960{x^2} = 2 \times 10(0.4 + x) \\
\Rightarrow 245{x^2} - 5x - 2 = 0 \\
\]
On solving the above quadratic equation, we have,
\[
\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\Rightarrow x = \dfrac{{5 \pm \sqrt {25 + 1960} }}{{490}} \\
\Rightarrow x = 0.1m \\
\]
The correct answer is option D.
Note: Many students do not consider the extra distance travelled by the mass x in the described problem. This means that the mass stops in air at the top end of the spring and through air the energy is transferred somehow in the spring and then it gets compressed. This is completely absurd.Always create a physical situation in your mind while solving such problems.
\[mg(h + x) = \dfrac{1}{2}k{x^2}\]
Complete step by step answer:
Let the compression in the spring be x. The point where the upper end of spring comes along with the mass of 2 kg is the level of zero gravitational potential.
Two of the most fundamental laws applicable in the whole universe are the Conservation of Energy and Conservation of Momentum. Here the conservation of Energy is applicable. The mass stored with some gravitational energy comes to the point of zero potential which is defined above. The gravitational potential is converted and stored as elastic potential energy in the spring. The spring is compressed by a length x.
The elastic potential energy of a spring is given as:
\[\Rightarrow {U_1} = \dfrac{1}{2}k{x^2}\]
The mass moves a distance of 40 cm along with a distance x. This is the same distance the spring compresses as the mass comes down along with the top end of the spring. So, the gravitational potential energy stored in the mass is:
\[\Rightarrow {U_2} = mg(h + x)\]
Equating the two equations and substituting the data from the question (let g=10m/s.s for simplicity), we have,
\[
\Rightarrow \dfrac{1}{2}1960{x^2} = 2 \times 10(0.4 + x) \\
\Rightarrow 245{x^2} - 5x - 2 = 0 \\
\]
On solving the above quadratic equation, we have,
\[
\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\Rightarrow x = \dfrac{{5 \pm \sqrt {25 + 1960} }}{{490}} \\
\Rightarrow x = 0.1m \\
\]
The correct answer is option D.
Note: Many students do not consider the extra distance travelled by the mass x in the described problem. This means that the mass stops in air at the top end of the spring and through air the energy is transferred somehow in the spring and then it gets compressed. This is completely absurd.Always create a physical situation in your mind while solving such problems.
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