# A Simple Explanation of the Law of Conservation of Mechanical Energy

The principle of conservation of mechanical energy is a fundamental law in physics and forms the foundation of various other laws and interrelated concepts. Let us know more about this concept.

Kundan Pandey

Last Updated: Jul 16, 2017

Principle

According to this law,

*in an isolated system, i.e., in the absence of non-conservative forces like friction, the initial total energy of the system remains constant.*Simply stated, the total mechanical energy of a system is always constant in case of absence of non-conservative forces. For instance, if a ball is rolled down a frictionless roller coaster, the initial and final energies will remain constant. Conservative forces are those that don't depend on the path taken by an object.
Equation

The quantitative relationship between work and energy is stated by the mechanical energy equation.

*U*, where,_{T}= K_{i}+ P_{i}+ W_{ext}= K_{f}+ P_{f}*U*= Total mechanical energy_{T}*K*= Initial kinetic energy_{i}*K*= Final kinetic energy_{f}*P*= Initial potential energy_{i}*P*= Final potential energy_{f}*W*= External work done_{ext}_{ext}is considered. In the absence of such forces, W

_{ext}= 0, so the mechanical energy conservation equation takes the form:

*U*

_{T}= K_{i}+ P_{i}= K_{f}+ P_{f}
Mathematical Example

Let us consider a mathematical problem that involves the use of this law in finding the values of unknown quantities.

*Question:*A 20 g stone is put in a sling shot with a spring constant of 100 N/m and it is stretched back to 0.7 m. Determine the maximum velocity that the stone will acquire and the speed at which it is shot straight up.*Solution:*In this problem, we ignore the air resistance and heat effects that are present while operating the sling shot. This makes the external work done zero, which means we can easily apply the law of conservation of mechanical energy.E

_{i}= K

_{i}+ Gravitational potential energy (mgh) + spring force (½ kx

^{2}).

Here,

K

_{i}= (0.5 mv

^{2}) = (0.5)m (0)

^{2}= 0 (Since v = 0 initially)

Gravitational potential energy = mg(0) = 0 (since h = 0 initially)

Spring force = ½ kx

^{2}= (0.5)(100)(0.7)

^{2}= 24.5 J = E

_{i}

Once out of the sling shot, the stone gains some maximum velocity before it reaches some altitude.

E

_{f}= 0.5 mv

^{2}+ mgh + ½ kx

^{2}= (0.5)(0.02)(v)

^{2}+ mg(0) + (0.5)k(0)

^{2}= 0.001v

^{2}

_{i}= E

_{f},

24.5 J = 0.001v

^{2}= 24,500 = v

^{2}. Therefore, v = 156.1 m/s (approximate value)

At the highest point, the velocity of stone is zero.

Therefore, E

_{f}= 24.5 J = 0.5 mv

^{2}+ mgh + ½ kx

^{2}

24.5 J = 0.5mv(0)

^{2}+ mgh + 1/2k(0)

^{2}= 24.5 J = (0.02)(9.8 N/Kg)h

= 125 m.

*Answer:*Velocity attained = 156.1 m/s and height attained = 125 m