Work Energy Theorem, Conservation Of Mechanical Energy (Board Level)

Energy

Energy of a body is its capacity of doing work. In mechanics a body is capable of doing work under two circumstances: (i) when it has motion and (ii) when it is situated in a field or when it is strained. When a bullet strikes a target it penetrates through the target and work is done against the resisting force offered by the target.

This energy is called kinetic energy of the body and is measured by the work it can do in being brought to rest.

Definition of kinetic energy:The capacity of doing work by virtue of motion is called kinetic energy and is measured by the total work that the body particle can do before coming to rest.

When a body is placed in a field, it can do work due to its position in the field. Such energy is called potential energy and is equal to the work done by an agent in placing it in the present position after bringing it from a position of zero field i.e., from infinity.

Due to position in earth’s gravitational field a body possesses energy called gravitational potential energy. Similarly, we have magnetic potential energy, electric potential energy. A compressed spring is capable of doing work due to strain produced in it. This is called elastic potential energy. The unit of energy is obviously the same as work.

Definition of potential energy:The capacity of doing work by virtue of position in a field or configuration is called potential energy and is measured by the work done by an agent in placing it in the present position after bringing it from infinity or the work done by the agent in producing the present configuration.

Units of Energy: The basic unit of energy is ‘joule’. In atomic and nuclear physics this is a too big unit. Here the convenient unit is ‘electron-volt’ (eV). This is defined as the energy gained by an electron when it falls through a potential difference of 1 volt. 1 eV = 1.6 × 10^–19 J.

1 million electron volt (1 MeV) = 10⁶ eV = 1.6 × 10^–13 joule.

When energy to be expressed is very large, joule is found inconvenient. There we use bigger units, viz. watt-hour and kilowatt-hour.

1 watt-hour = 3600 joule and 1 killowatt-hour (kWH) = 3600 × 1000 joule = 3.6 megajoule (MJ).

Kinetic Energy and the Work-Energy theorem

Let a particle of mass m move with velocity v. To calculate its kinetic energy we have to see what amount work it can do in being brought to rest by an external opposing force. Let F be a constant opposing force applied on the particle and let it be brought to rest within s.

from third equation v_f²– v_i²= 2as where a = retardation produced by F, Put v_f= 0 , v_i = v

we get     \displaystyle s=\frac{{{{v}^{2}}}}{{2a}}

Now               KE_f– KE_i =   Work done = F.s.cos(180)                    ( but KE_f = 0 and let KE_i = K)

-K  = – ma\times \frac{{{{v}^{2}}}}{{2a}}                                                (QF = ma)

K= \frac{1}{2}m{{v}^{2}}.

Kinetic energy (K) of a particle of mass having velocity v=\frac{1}{2}m{{v}^{2}}. 

In terms of momentum, K=\frac{{{{p}^{2}}}}{{2m}}                                    (Qp = mv)            

Now we will show that the change in the kinetic energy of a particle is always equal to the work done on the particle by the force whether the force is constant or variable.

We have in general,

W=\int_{{{{x}_{1}}}}^{{{{x}_{2}}}}{{Fdx=}}\int_{{{{x}_{1}}}}^{{{{x}_{2}}}}{{m\frac{{dv}}{{dt}}}}\ dx

\left( {\because \ \ \ \ \ \ F=m\frac{{dv}}{{dt}}\text{ by Newton }\!\!’\!\!\text{ s law}} \right)

or        W=\int_{{{{x}_{1}}}}^{{{{x}_{2}}}}{{m\frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}}}\,dx

or        W=m\int_{{{{x}_{1}}}}^{{{{x}_{2}}}}{{\frac{{dv}}{{dx}}.v}}\,dx        

\left( {\because \ \ \ \ \ v=\frac{{dx}}{{dt}}} \right)

=\int_{{{{v}_{1}}}}^{{{{v}_{2}}}}{{mv\ dv}}=\frac{1}{2}mv_{2}^{2}-\frac{1}{2}mv_{1}^{2}

or        W = K₂ – K₁ = ΔK.

Thus the work done on a particle by a force (constant or variable) is always equal to the change in the kinetic energy of the particle. This is known as the work-energy theorem.

Illustration

A block weighing 5 kg slides on a horizontal frictionless table with a speed of 1.2 ms^–1. It is brought to rest in compressing a spring in its path. By how much is the spring compressed if its force constant is 3.5 Newton per metre?

Solution

By the ‘work energy’ theorem, Work done = change in kinetic energy.

Here work done =\frac{1}{2}k{{x}^{2}}

  \frac{1}{2}k{{x}^{2}}=\frac{1}{2}m{{v}^{2}}, or  x=\sqrt{{\frac{m}{k}}}.\ v=\sqrt{{\frac{5}{{3.5}}}}.1.2

or,          x = 1.43 metre. Ans.

Potential Energy

Potential energy is energy due to position. If a body is in a position such that if it were released it would begin to move, it has potential energy.

There are two common forms of potential energy, gravitational and elastic.

(a) Gravitational Potential Energy, it is a property of height.