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Work Power Energy : Theory & Examples 0/3

Lecture1.1

Lecture1.2

Lecture1.3

Work
Work is a method by which energy is transferred from one body to another body by the application of force. In above example energy is being transferred from human to block.
Positive Work
When energy is transferred (given) to the object by the application of force than work done by that force will be positive i.e. in above example work done by human is positive.
Negative work
When energy is extracted from an object by the application of force than work done by that force on the object will be negative. In above example if human was pushing instead of pulling in opposite direction of movement than work done by human would have been negative.
Zero work
When there is no exchange of energy between the two objects due to a force than work done by that force will be zero.
Mathematical Definition of work
F is the force whose work is to be calculated.
dS is the displacement of point of application of force with respect to observer.
Work is a Frame dependent Quantity i.e. value of work depends on the observer as well.
Theta is the angle between the force and displacement vector.
(in this case we’re assuming that force applied and displacement are nonzero)
Let’s learn few concepts through Examples
Q1. State TRUE or FALSE
Statement: A man starts running on a rough horizontal surface and acquires some velocity without slipping than work done by frictional force on the man will be zero.
Answer: TRUE
Explanation:
As shown in figure, as there is no slipping, displacement of leg is zero while friction force is acting. Hence work done by friction force will be zero in this case.
Q2. State TRUE or FALSE
Statement: A man is climbing up the stair case. During this process work done by normal reaction on the man will be zero.
Explanation:
While climbing on the stair cases, while foot is in contact with stair case and normal reaction is acting, displacement of the foot in vertical direction is zero. Similarly the moment displacement happens i.e. foot goes into the air, there is no normal reaction acting on the foot as there is no contact between staircase and foot. Hence work done by normal reaction on man will be zero in this case.
Q3. During the jumping process on a horizontal surface, in case of vertical jump, work done by the normal reaction by the ground on the man will be?
Ans: This is the similar case as in Q2. Duration in which Normal reaction is acting on the man, displacement is zero and the moment displacement happens, normal reaction is zero. Hence work done by ground’s normal reaction on the man will be zero in this case as well.
Q4. A rubber ball is thrown towards a rigid wall. During the time velocity of the ball becomes zero from initial velocity, work done by the normal reaction of wall on the ball will be? (Positive/Negative/Zero) ?
Ans:
As shown in above figures in (a) ball has just come into contact with wall, with velocity V which is greater than zero. After some time ball’s velocity becomes zero. Now at the point of contact, Normal reaction is acting from the wall on the ball, but the displacement in this case is zero. Hence work done by wall on the ball due to normal reaction will be zero.
Q5. A rigid ball is thrown towards a rubber wall. During the time the velocity of ball becomes zero from initial velocity, work done by the normal reaction exerted by the wall on the ball will be? (Positive/Negative/Zero)?
Ans
Please note, in this case ball is rigid and wall is of rubber, hence there will not be any deformation in ball, but there will be some displacement in wall as it’s made of rubber. Hence ball will cover some distance before coming to rest while normal reaction from wall is working on it. In this case, ball is moving in the opposite direction of normal reaction and hence work done by wall on the ball due to normal reaction will be negative in this case.
Q6. What will be the work (positive/negative/zero) if both wall and ball are made of rubber in above case?
Ans
Negative. As in this case there will be some deformation in ball and some deformation in wall as well. Hence ball will travel some distance before coming to rest. Displacement will be in opposite direction of normal reaction; hence in this case as well work done by wall on the ball due to normal reaction will be zero.
Q7. State TRUE or FALSE
Statement: Work done by kinetic friction is always negative.
Ans: False
Explanation:
Let’s take example as shown in figure above. A force F1 has been applied on mass M1, it slides on mass M2, and because of that kinetic friction acts between two blocks. As surface below M2 is smooth, kinetic friction acting on Mass M2 will move it in positive direction. Hence work done by kinetic friction on mass M2 will be positive in this case.
Q8. State True or False
Statement: Work done by kinetic friction cannot be zero.
Ans: False
Explanation:
In above example, displacement of ground due to kinetic friction is zero. Hence work done by kinetic friction on ground is zero.
Q9. State True or False.
Statement: Work done by static friction is always zero.
Ans: False
Explanation:
In above figure, as max friction is more than applied force, there will be no sliding between M1 & M2. Hence M1 will move in positive direction (left direction, in the direction of force) due to static friction (as there is no sliding between M1 & M2, only tendency of sliding). Hence work done by static friction on Mass M1 is positive.
Now let’s do some Numerical Examples
Q10.
Find out work done by applied force on A & B, and work done by Normal reaction on A & B during the time block displaces 10 meters.
Ans:
Let’s draw Free body diagram,
Find out value of normal reaction by solving force balance equation on both blocks,
Now calculate, Work done by Force F on block A & B,
Here, as force F is not acting on block B, work done by F on block B will be equal to zero.
Calculate work done by Normal reaction on A & B,
Let’s find out Kinetic Energy of system,
Velocity of both blocks, initial velocity = 0, hence
K.E. of Block A,
K.E. of Block B,
K.E. of System,
2000 + 1000 = 3000 J
Things to notice here,
Work done by Force F has been stored in the form of kinetic energy in system.
Work done by Normal reaction on the system is zero (positive on block B and negative on block A).
Q11.
Find out work done by applied force on Block A (right) & work done by Tension on A & B (left) during the time blocks displaces by 10 meters.
Ans:
As usual, let’s first draw free body diagram,
Let’s write force balance equation to find out Tension force,
Note : String connecting two blocks is not massless here, hence there will different tension on the sides of string.
Hence,
Work done by F on A = 100 x 10 x cos 0 = 1000J
Work done by T1 on A = 60 x 10 x cos 180 = 600 J
Work done by T2 on B = 40 x 10 x cos 0 = 400 J
Q12.
Find out work done by the normal, gravity & friction during the time Block reaches at the bottom.
Ans
Let’s Draw FBD first,
First of all, let’s check whether this block will slide or not on above given inclined plane,
Friction force acting on the block (assuming that block is sliding)
Apply force balance parallel to inclined surface,
It’ given that,
Hence value of acceleration is nonzero, block will slide on the inclined plane.
Now, calculate work done by specific forces,
Work done by gravity,
As we know that gravity is working in vertically downward direction, and block has moved h distance in vertical direction, hence work done by gravity on the block will be = Mgh
Work done by normal reaction,
There is no displacement in the direction perpendicular to the inclined surface, i.e in the direction of normal reaction; work done by normal reaction on the block is zero.
Work done by friction,
f x displacement in the direction of force
Q13. Find out work done by given Force vector on moving a particle from given initial position to final position.
Initial position = (0,0), Final position = (2,3)
Ans:
We have given a Force vector, what is required to calculate work done is the displacement vector.
Let’s calculate displacement vector,
Now, work done by force is nothing but dot product of force and displacement vector,
Calculating Potential Energy of a Spring
Consider spring shown in above diagram. A force is applied which is gradually increases so that extension of spring becomes y. i.e. length of spring becomes L + y where L is the natural length of spring.
Potential energy of spring is nothing but energy transferred to spring or net work done on the spring. Let’s calculate work done by Force F and the wall on the spring in above case.
Work done by wall on the spring is zero, as there is no displacement of the point where tension force is acting on the spring due to wall.
Now, calculate work done by Force F,
In this case value of Force F is changing as length of spring changes, hence we have to add work done at each dx extension. So let’s take a moment when –
Extension in spring = x
Elongation in spring = dx
Force on the spring at this moment, F = kx
Hence, energy in the spring or Potential energy of spring = ½ ky^2
Spring with spring constant k, with extension x, will have potential energy U = ½ Kx^2
Work done on a spring in terms of potential energy
Work done on a spring = Change in Potential energy of spring
Work done on a spring = Potential Energy final – Potential Energy initial
W = Uf – Ui
So if spring in above example is extended from y to 2y,
Work done on spring = Uf – Ui
Q14.
Find out work done on spring in above diagram, assuming that initially spring was in natural length i.e. extension was zero.
Ans:
Work done on spring =Uf – Ui
Tota extension in spring = x + 2x = 3x
Hence, work done is
Let’s find out work done by both persons, assuming left person is A, & right one is B.
Assuming that spring is massless, in that case tension on both ends of spring will be equal. Hence both persons are putting equal amount of force on spring. Thus work done by both persons will be in the ratio of displacement done by them.
So,
Wb = 2 Wa
Q15. Find out work done by Force F in below force displacement graph.
Ans:
In above graph, a particle is moving in negative x direction, from initial position of x = 5 to final position x = 0 and a positive force F is working, which is constant and equal to 10N.
Now, it’s a simplified problem,
W = F. displacement
= – 10 x 5
= – 50 J
Work Energy Theorem
Work done by all forces = change in Kinetic Energy
Work done by internal forces may or may not be zero. It depends on the compression or extension with in the body due to internal forces.
In case of a rigid body there is no compression or extension within the body, so force rigid bodies work done by internal forces is always zero.
Hence, for rigid bodies,
Work done by external forces = change in Kinetic energy
Q16.
Find out work done by Air resistance on the block.
Ans:
Let’s apply Work energy theorem,
Work done by all forces on block = change in kinetic energy of block
Initial velocity of block = 0, final velocity as given in diagram
As, block is a rigid body, work done by internal forces will be equal to zero, hence
Few important facts about Kinetic Energy
 Kinetic energy is a frame dependent quantity because velocity is a frame dependent quantity
 Kinetic energy can never be negative, K.E > 0
Kinetic Energy in the form of Momentum
Where P is momentum of particle.
Q17. State True of False.
Statement : If the momentum of system of two particles is zero, then the Kinetic energy of the system may not be zero.
Ans : TRUE
Explanation:
As shown in above case, momentum of the system is zero, as momentum due to one particle is opposite and equal to that of another particle. But in kinetic energy, everything is positive and hence it adds up instead of cancelling out.
Work done by Tension force on System
Assuming strings are massless and inextensible, string can’t have kinetic energy and potential energy, So it always transfer energy from one object to another object completely. And hence net work done by tension force on the system is always zero in such cases.
Q18. Find out relation in velocity of block A & B in below pulley system.
Ans:
As we have learnt recently that work done by tension force on the system is always zero. We’ll use this principle to solve above problem.
Let’s take current moment and calculate work done by tension force on the system for displacement happened in “dt” time. (dt time is very small time).
Work done by Pseudo force
Let’s take an example. Two person A & B are there. Person A is on ground and is at rest while person B is standing and holding the wall of truck, which is moving with acceleration = 10 m/s^2 from rest.
For person A, when truck will start moving, it appears that truck is moving with acceleration = 10 and person standing in the truck as well. While block of 100 kg will appear to be at rest.
Now what will person B sees,
For person B, truck is at rest, and block of 100 kg is moving towards him with an acceleration of 10 m/s^2. Person B will see the block moving but not any force being applied on the block. In such cases we assume a pseudo force (because in actual there isn’t any force acting on the block) acting on the block.
Please note, pseudo force will come into action only if reference frame is moving with some acceleration.
How to apply work energy theorem with respect to accelerated reference frame
To explain the change in energy in accelerated reference frame, work of pseudo force is also considered. Without considering pseudo force work energy theorem will be incorrect w.r.t. accelerated reference frame.
So,
Value of pseudo force will be equal to Mass x Acceleration, and in opposite direction of acceleration of reference frame.
Q19. Find out the maximum compression in the spring in below springblock system.
Ans:
We know that initial velocity is zero and final velocity will also be zero. (because velocity of block will be zero at the moment when compression in the spring is maximum)
Psuedo force in this case = Ma (Mass of the block x acceleration)
Let’s apply work – energy theorem,