You must remember the following points to make you strong in answering multiple choice questions involving *work, energy and power*:

**(1)**Work is done by a force if the point of application of the force is

*displaced*. If the force vector

**F**is constant and makes an angle

*θ*with the displacement vector

**r**, the work done (

*W*)

*is given by*

*W = Fr cos* *θ*

In vector notation this is given by

*W* = **F.r**

Thus work is a scalar quantity given by the *scalar product* of force and displacement.

If the force is not constant and is a function of the displacement (as for instance, in compressing or elongating a spring), the work done is given by* W* = ∫**F.dr** where **dr **is a small displacement for which the force **F** can be taken to be constant. The integration is to be carried out over the entire displacement. The work done by you in producing an elongation ‘x_{1}’* *in a spring (which is not deformed initially) of force constant ‘k’ is given by

*W* = _{0}∫^{x}^{1}^{ }**F.dx **= _{0}∫^{x}^{1}^{ }kxdx = ½ kx_{1}^{2}

[Note that the force with which you have to pull the spring to produce an extension ‘x’ is kx]

The work done in producing an extension ‘x’ in a spring is ½ kx^{2}.

*x*

_{1}to

*x*

_{2}the work required is ½ k(x

_{2}

^{2 }– x

_{1}

^{2}).

In the case of a spring the force-elongation (displacement) curve is a straight line as shown in the figure. The work done in producing an elongation ‘x’ in a spring is the area under the curve (which is the area of the triangle OAB) and is equal to ½ Fx = ½ (kx)x = ½ kx^{2}.

**(2) **A force is** conservative** if the work done by the force in moving an object depends only on the initial and final positions of the object and is independent of the path followed between these positions. The total work done by a conservative force on an object is zero when it moves round any closed path and returns to the initial position. *Gravtational force* and *electrostatic force* are examples of conservative forces.

If the work done by a force on an object moving between two positions depends on the path taken, the force is called **non-conservative **force. *Friction* and *visous force* are non-conservative forces.

**(3)** **Kinetic energy** (*K*) of a body of mass *m* moving with velocity *v* is given by

*K* = ½ *mv*^{2}

Since momentum, **p** = m**v, **we have

*K* = *p*^{2}/2*m*

[Remember this useful expression].

**(4)** **Work-energy theorem** states that the work done by a force acting on a body is equal to the change in the kinetic energy of the body.** **If *W *is the work done and *K _{i}* and

*K*are the initial and final kinetic energies respectively, we have

_{f} *W = K _{f} * –

*K*

_{i}**(5) The gravitational potential energy **of a body of mass** ***m* at a height *h* near the surface of the earth is ** mgh** where

*g*is the acceleration due to gravity at the place. Note that this energy is with respect to the reference level used for measuring the height and the value of

*h*is

*negligible*compared to the radius of the earth.

The change in the gravitational potential energy (∆*U*_{g}) of a mass *m *raised through a small height *h* can therefore be written as

∆*U*_{g} = *mgh*

You will find more details on gravitational potential energy in the posts dated 9^{th}, 12^{th} and 17^{th} May 2008 which you can access by clicking on the label ‘gravitation’ below this post.

**(6) **When a spring is compressed or elongated, the work done on the spring is stored as elastic potential energy in the spring. The elastic potential energy in the spring which is elongated or compressed through a distance ‘x’ is ½ kx^{2}.

**(7) **Law of conservation of energy states that energy can neither be created nor destroyed but can only be transferred from one form to another.

You should understand that the production of energy by annihilating mass in nuclear reactions does not violate the law of conservation of energy since mass itself is to be treated as a cocentrated form of energy in accordance with Einstein’s mass-energy relation, *E = mc*^{2}.

**(8) Power **is the time rate at** **which work is done or energy is transferred.

The average power *P _{av} = W/t* where

*W*is the total work done in a time

*t.*

Instantaneous power *P* = d*W/*d*t* where d*W* is the work done in a very small time d*t* at the instant *t*.

**(9) **The instantaneous power can be expressed as the scalar product of the force vector **F **and the velocity vector **v** as

*P = ***F.v*** *

This is easily obtained since d*W = ***F.dr** so that *P* = d*W/*d*t* = **F.**(**dr**/d*t*) = **F.v**

**(10) Elastic collision** is one in which momentum and *kinetic energy* are conserved. **Inelastic collision** is one in which *kinetic energy is not conserved*, but momentum is conserved.

You should note that momentum is conserved in elastic as well as inelastic collisions; but kinetic energy is conserved in the case of elastic collisions only.

**the two colliding bodies**

*completely inelastic collision**move together*after the collision.

If the mass *m*_{1} moves with velocity *v _{1i} *in the positive x-direction and suffers a

*completely inelastic*head on collision (collision in one dimension) with the mass

*m*

_{2}at rest (fig.), the common velocity (

*v*

_{f}) with which the two masses will move is given by

*m*_{1}*v*_{1i}* = *(*m*_{1 }+ *m*_{2}) *v*_{f}, on applying the law of conservation of momentum.

Therefore, *v _{f}* =

*m*

_{1}

*v*

_{1i }

*/*(

*m*

_{1 }+

*m*

_{2})

If the collision is *elastic* in the case of the above masses, we will have different

final velocities *v*_{1f }and *v*_{2f} for the two bodies after the collision, as given by the momentum conservationlaw:

*m*_{1}*v*_{1i}* = m*_{1}* v*_{1f }+ *m*_{2} *v*_{2f} -------(i)

Since there are two unknowns (*v*_{1f }and *v*_{2f}) we require one more equation to solve for the unknowns and we have the kinetic energy equation,

½ *m*_{1}*v*_{1i}^{2}* =*½ * m*_{1}* v*_{1f }^{2}+ ½ *m*_{2} *v*_{2f} ^{2}

Or, *m*_{1}*v*_{1i}^{2}* = m*_{1}* v*_{1f }^{2}+ *m*_{2} *v*_{2f} ^{2}-------(ii)

Equations (i) and (ii) give

*v*_{1f} = (*m*_{1}–* m*_{2})* v*_{1i} */*(*m*_{1 }+ *m*_{2}) and

*v*_{2f} = 2* m*_{1}*v*_{1i} */*(*m*_{1 }+ *m*_{2})

__If the the two bodies are of the same mass, the colliding mass m_{1} will come to rest and will hand over its velocity to the other mass m_{2} which was initially at rest. If m_{2}>> m_{1}, the velocity of m_{1} will get reversed and m_{2} will continue to remain at rest.__

If *m*_{2} also has an initial velocity *v*_{2i} you can incorporate its initial momentum and kinetic energy in the above equations and solve for the final velocities of the bodies. You will then get

*v*_{1f} = [2* m*_{2}*v*_{2i} + (*m*_{1}–* m*_{2})* v*_{1i}] */*(*m*_{1 }+ *m*_{2}) and

*v*_{2f} = [2* m*_{1}*v*_{1i} + (*m*_{2}–* m*_{1})* v*_{2i}] */*(*m*_{1 }+ *m*_{2})

Try to derive these equations as an exercise.

a blog on physics! I haven't seen that before! I never liked physics like I should have when I was in college. I don't know why :(

ReplyDeleteJust dropping by! Keep up the good work! I bet there are many people especially physics students finding your site very useful!