Question

A toy gun uses a spring of force constant $k$. When charged before being triggered in the upward direction the spring is compressed by $x$. If the mass of the shot is $m$, on being triggered, it will go upto a height of :
A. \[\dfrac{{{\text{k}}{{\text{x}}^2}}}{{2{\text{mg}}}}\]
B. \[\dfrac{{{\text{k}}{{\text{x}}^2}}}{{{\text{mg}}}}\]
C. \[\dfrac{{{{\text{k}}^2}{{\text{x}}^2}}}{{{\text{mg}}}}\]
D. \[\dfrac{{{{\text{x}}^2}}}{{3\;{\text{kg}}}}\]

Answer 1

Hint:The spring factor is the force acting on the unit extension formed. The stress exerted by both springs is the same when two springs are paired in sequence. Specific extension is the sum of total extension (X).

Complete step by step answer:
Energy cannot be created or destroyed, according to the law of conservation of energy. It can only be transferred from one source of energy to another. Unless energy is added from the outside, a device still has the same amount of energy.

This is particularly perplexing in the case of non-conservative forces, in which energy is transferred from mechanical to thermal but the total energy remains constant. Energy can only be used by converting it from one type to another.

Potential energy accumulated as a result of deformation of an elastic material, such as the stretching of a spring, is known as elastic potential energy. It is proportional to the effort required to extend the spring, which is determined by the spring constant k as well as the length of the spring.

According to the law of conservation of energy, elastic potential energy accumulated in the spring (\[\dfrac{1}{2}k{x^2}\]) = gravitational potential energy obtained by the projectile (mgh)
\[\dfrac{1}{2}k{x^2} = mgh\]
\[\therefore h = \dfrac{{k{x^2}}}{{2mg}}\]

Hence, option A is correct.

Note:Energy that is contained – or conserved – in an object or material is referred to as potential energy. The accumulated energy is determined by the object's or substance's location, structure, or condition. It can be thought of as capacity with the 'potential' to do work.