Physics

Session

Work, Power and Energy - 1

Session Objectives

Session Objective

1. Work done by constant force

2. Work done by variable force

3. Kinetic Energy

4. Work-Energy Theorem

5. Conservative and non-conservative forces

6. Potential Energy and Total Mechanical Energy

F

cosF

F

cosF

F

cosF

F

cosF

Displacement S

Component of force in the direction of displacement = Fcos

Work done = (FcosS =FS cos

W =F.S

Work done by constant Force

x

x

xF

Total work done is sum of all terms from xi to xf

xfx

xi

W = F Δx

ix fx

xF

Work by Varying Force

x

xF

ix fx

xfx

x 0 xi

xf

xxi

W lim F x

W F dx

Work by Varying Force

Conservative Forces

Non-Conservative Forces

Force of gravity & spring ForceForce of gravity & spring Force

Frictional Force& Viscous forceFrictional Force& Viscous force

Work done on a particle between any two points is independent of the path taken by the particle.

Work done on a particle between any two points depends on the path taken by the particle.

Conservative Forces : Work done will be same

Non-Conservative Force : Work done will be different

Illustration of principle

21KE = mv

2

Nature : scalarUnit : joules(J)

Energy associated with the motion of a body.

Kinetic Energy

Total work done by all the forces acting on a body is equal to the change in its kinetic energy.

total fiW =KE - KE

c nc ext fiW +W +W =KE - KE

Conservative forces (Wc)

Non-Conservative forces (Wnc)

External forces (Wext)

Work done by

Work Energy Theorem

xf

c x i fxi

W = F dx = -ΔU =U - U

xf

f x ixi

U = - F dx +UUi can be assigned any value as only U is important

Conservative Forces & Potential Energy

Work done by a conservative force equals the decrease in the potential energy.

Force varies with position. Fs=-kx[k :Force constant]

0x x

kxFs

x

kxFs 0Fs

Work done in compression/extension of a spring by x

21W = kx

2= PE stored in the spring

Conservative Force: Spring force

Class Test

Class Exercise - 1

Force acting on a body is

(a) dependent on the reference frame

(b) independent of reference frame

(c) dependent on the magnitude of velocity

(d) None of these

A specific force is external in origin and so is independent of reference frame.

Solution :

Hence answer is (d).

Class Exercise - 2

<<<<<<<<<<<<<<1r 2 i 3 j

<<<<<<<<<<<<<<2r 3 i 2 j

A particle moves from a point

to another point during which a

certain force acts on it. The work

done by the force on the particle during displacement is:

F 5 i 5 j

(a) 20 J (b) 25 J (c) zero (d) 18 J

<<<<<<<<<<<<<<<<<<<<<<<<<< <<2 1Displacement d r – r i – j

F d 5 i j i – j 0

Solution :

Hence answer is (c)

Class Exercise - 3A force F = (a + bx) acts on a particle in the x-direction where a and b are constants. The work done by this force during a displacement from x = 0 to x = d is

(a) zero (b)

(c) 2a + bd (d) (a + 2bd)d

1a bd d

2

Force is variable.

dd d 22

0 0 0

bx bdW Fdx (a bx)dx ax ad

2 2

Solution :

Hence answer is (b)

Class Exercise - 4

A block starts from a point A, goes along a curvilinear path on a rough surface and comes back to the same point A. The work done by friction during the motion is:

(a) positive (b) negative

(c) zero (d) any of these

Friction always acts opposite to the displacement.

Solution :

Hence answer is (b).

Class Exercise - 5

The work done by all forces on a system equals the change in

(a) total energy

(b) kinetic energy

(c) potential energy

(d) None of these

Statement characterizes kinetic energy.

Solution :

Hence answer is (b).

Class Exercise - 6

A small block of mass m is kept on a rough inclined plane of inclination fixed in an elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be

(a) zero (b) mgvtcos2

(c) mgvtsin2 (d) mgvtsin2

Solution

f = mg sin as block does not slide.

Displacement d = vt

Angle between d and f = (90 – )

W = fd cos(90° – ) = mgvt sin2

Sin mg

mg

vt

f

Class Exercise - 7

A force (where K is a

positive constant) acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force on the particle is:

F –K y i x j

(a) –2Ka2 (b) 2Ka2

(c) –Ka2 (d) Ka2

Solution

First path: y = 0

F x j (perp. to displacement) W 0

Second path: x = a

F ( K) y i a j

Displacement along x = 0

a

2y

0

W F dy –Ka

y

x

a

O 1

2

a

Class Exercise - 8

Under the action of a force, a 2 kg body moves such that its position

x as a function of time is given by3t

x3

where x is in metre and t in seconds. The work done by the force in the first two seconds is:

(a) 1,600 J (b) 160 J (c) 16 J (d) 1.6 J

Solution

3 22

2t dx d x

x , t , F m 2mt3 dt dt

Hence answer is (c)

x 2 2 4

2 3

0 0 0

mtW Fdx 2mt t dt 2m t dt 16 J

2

Class Exercise - 9There is a hemispherical bowl of radius R. A block of mass m slides from the rim of the bowl to the bottom. The velocity of the block at the bottom will be:

(a) Rg (b) 2Rg

(c) 2 Rg (d) Rg

Solution

Hence answer is (b)

PE = mgR

21KE mv

2

21mv mgR

2 v 2Rg

R

PE = mgR

KE = mv212

Class Exercise - 10

A block of mass 250 g slides down an incline of inclination 37° with uniform speed. Find the work done against the friction as the block slides down through 1.0 m.

4cos37 , g 10 m/s

5

(a) 15 J (b) 150 J (c) 1.5 J (d) 1500 J

Solution

As the block slides with uniform speed, net force along the incline is zero.

Hence answer is (c)

Mg sin37° = N

Work done by gravity = Work done against friction = Mg sin37° x s

4 3cos37 , sin37°

5 5

3

W 0.25 10 15

= 1.5 J

Mg cos37oMg sin37o

mg

N F=N

37o

Thank you