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Work Power Energy : Theory & Examples 0/3
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Work

work example

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

$dW =\overrightarrow {F} \cdot \overrightarrow {dS}$

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.

$dW =| \overrightarrow {F} | \cdot | \overrightarrow {dS} | \cos \theta$

Theta is the angle between the force and displacement vector.

$dw > 0 \quad \Rightarrow \quad \cos \theta > 0 \quad \Rightarrow \quad 0 \le \theta < 90^{0}$

force at angle

$dw < 0 \quad \Rightarrow \quad \cos \theta < 0 \quad \Rightarrow \quad 90^{0} < \theta \le 180^{0}$

force at 180 degree

$dw = 0 \quad \Rightarrow \quad \cos \theta = 0 \quad \Rightarrow \quad \theta = 0$

(in this case we’re assuming that force applied and displacement are non-zero)

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:

work done on leg

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:

work done in climbing stair case

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:

rubber ball and rigid wall

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

work done rigid ball and rigid wall

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:

work done by kinetic friction

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:

zero work by kinetic friction

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:

zero work by static friction

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.

work done on two blocks

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,

work done on two blocks fbd

Find out value of normal reaction by solving force balance equation on both blocks,

$N = 50 \cdot a \newline F-N = 100 \cdot a \newline F = 300 = 150 \cdot a \quad \Rightarrow \quad a = 2 m/s^2$

Now calculate, Work done by Force F on block A & B,

$(W_F)_{on \quad A} = 300 \cdot 10 = 3000J \newline (W_F)_{on \quad B} = 0$

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,

$(W_N)_{on \quad A} = 100 \cdot 10 \cdot \cos {180^0}= -1000J \newline (W_N)_{on \quad B} = 100 \cdot 10 \cdot \cos {0^0} = 1000J$

Let’s find out Kinetic Energy of system,

Velocity of both blocks, initial velocity = 0, hence

$V^2 = 2as = 2 \cdot 2 \cdot 10 = \quad 40 \quad m^2/s^2$

K.E. of Block A,

$\frac{1}{2} MV^2 = 2000J$

K.E. of Block B,

$\frac{1}{2} MV^2 = 1000J$

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.

work done by tension

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.

$100 - T_1 = 20 \cdot a$

$T_1 - T_2 = 10 \cdot a$

$T_2 = 20 \cdot a$

$\Rightarrow 100 = 50 \cdot a$

Hence,

$T_1 = 60N \newline T_2 = 40 N$

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.

work done slope

Find out work done by the normal, gravity & friction during the time Block reaches at the bottom.

Ans

Let’s Draw FBD first,

work done slope solution

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)

$f = \mu Mg \cos \theta$

Apply force balance parallel to inclined surface,

$Mg \sin \theta - f = M \cdot a \newline Mg \sin \theta - \mu Mg \cos \theta = M \cdot a \newline g (\tan \theta - \mu) = M\cdot a$

It’ given that,

$\tan \theta > \mu$

Hence value of acceleration is non-zero, 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

$W_f = - \mu Mg \cos \theta \frac{h}{\sin \theta} \newline -\mu Mg h \cot \theta$

Q13. Find out work done by given Force vector on moving a particle from given initial position to final position.

$\overrightarrow{F} = 2 \hat i + 3 \hat j$

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,

$(2-0) \hat i + (3-0) \hat j = 2 \hat i + 3 \hat j$

Now, work done by force is nothing but dot product of force and displacement vector,

$W = \overrightarrow F \cdot \triangle \overrightarrow S \newline = ( 2 \hat i + 3 \hat j) \cdot ( 2 \hat i + 3 \hat j) \newline = 4 + 9 \newline= 13 units$

Calculating Potential Energy of a Spring

potential energy 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

work done by spring

$dW = Fdx \newline\int dW = \int kx dx \newline \left[ \frac{kx^2}{2} \right]^y_0= \frac{ky^2}{2}$

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

$\frac{1}{2}k (2y)^2 - \frac{1}{2} k y^2 = \frac{3}{2} k y^2$

Q14.

two men work on spring

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

$W = \frac{1}{2} k (3x)^2 = \frac{9x^2k}{2}$

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

$W_b = 2W_a \newline W = W_a + W_b = 3W_a \newline \frac{9kx^2}{2} = 3W_a \newline W_a = \frac{3kx^2}{2} \newline W_b = 3kx^2$

Q15. Find out work done by Force F in below force displacement graph.

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

$W_{all} = \triangle K$

Work done by all forces = change in Kinetic Energy

$W_{ext} + W_{int} = \triangle K$

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.

work done by air resistance

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

$W_g + W_{air} + W_{int} = \triangle K \newline mgh + W_{air} + 0 = \frac{1}{2} m ({\sqrt {gh}})^2 - 0 \newline W_{air} = \frac{-mgh}{2}$

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

$k = \frac{1}{2} mv^2 \newline k = \frac{1}{2} m \overrightarrow V \cdot \overrightarrow V = \frac{1}{2m} m \overrightarrow V \cdot m \overrightarrow V \newline k = \frac{P^2}{2m}$

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:

momentum of two particle system

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.

$k=\quad \frac { P_1^{ 2 } }{ 2m } +\quad \frac { P_2^{ 2 } }{ 2m }$

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.

$(W_t)_{sys} = 0$

Q18. Find out relation in velocity of block A & B in below pulley system.

find velocity in 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 energy theorem in pulley system

$(W_t)_{sys} = 0 \newline (W_t)_A + (W_t)_{pulley} + (W_t)_{roof} + (W_t)_{roof} + (W_t)_{string} + (W_t)_B = 0 \newline \frac{T}{2} V_A dt + 0 + 0 + 0 - 3T (V_bdt) = 0 \newline V_A = 6V_B$

Work done by Pseudo force

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,

$W_{all} + W_{psf} = \triangle K$

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 spring-block system.

work done on spring system pseudo force

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,

$W_{all} + W_{psf} = \triangle K = 0 \newline W_g + W_N + W_{spring} + W_{psf} = 0 \newline -\frac{1}{2}kx^2 + M \cdot a \cdot x = 0 \newline Max = \frac{1}{2}kx^2 \newline x = \frac{2Ma}{k}$

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Physics

Work Power Energy : Theory & Examples 0/3

Work

Positive Work

Negative work

Zero work

Mathematical Definition of work

Let’s learn few concepts through Examples

Now let’s do some Numerical Examples

Calculating Potential Energy of a Spring

Work done on a spring in terms of potential energy

Work Energy Theorem

Few important facts about Kinetic Energy

Kinetic Energy in the form of Momentum

Work done by Tension force on System

Work done by Pseudo force

How to apply work energy theorem with respect to accelerated reference frame

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