Transcript 12. Work Power & Energy

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
Work done by constant Force
F

F
F
F



F cos
F cos
 F cos 
F cos 
Displacement S
Component of force in the direction of displacement
= Fcos
Work done = (Fcos) S =FS cos
W = F.S
Work by Varying Force
Fx
Fx
Total work done is sum
of all terms from xi to xf
xf
x
xi
x
xf
W =  Fx Δx
xi
Work by Varying Force
Fx
xf
W  lim  Fx x
x 0 x
i
xf
x
xi
xf
W   Fxdx
xi
Conservative Forces
Work done on a particle between
any two points is independent of
the path taken by the particle.
Force of gravity & spring Force
Non-Conservative Forces
Work done on a particle between
any two points depends on the
path taken by the particle.
Frictional Force
& Viscous force
Illustration of principle
Conservative Forces :
Work done will be
same
Non-Conservative Force :
Work done will be different
Kinetic Energy
Energy associated with the motion
of a body.
1
KE = mv2
2
Nature : scalar
Unit : joules(J)
Work Energy Theorem
Total work done by all the forces
acting on a body is equal to the
change in its kinetic energy.
Wtotal = KEf - KEi
Work done by
Conservative forces (Wc)
Non-Conservative forces (Wnc)
External forces (Wext)
Wc + Wnc + Wext = KEf - KEi
Conservative Forces & Potential
Energy
Work done by a conservative
force equals the decrease in
the potential energy.
xf
Wc =  Fxdx = -ΔU = Ui - Uf
xi
xf
Uf = -  Fxdx + Ui
xi
Ui can be assigned
any value as only U
is important
Conservative Force: Spring force
Force varies with position.
Fs=-kx
[k :Force constant]
Fs  0 Fs  kx
Fs  kx
x x 0
x
Work done in compression/extension of a spring by x
1
W = kx2
= PE stored in the spring
2
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
Solution :
A specific force is external in origin and
so is independent of reference frame.
Hence answer is (d).
Class Exercise - 2


A particle moves from a point r 1  2 i  3 j


to another point r2  3 i  2 j during which a


F  5 i  5 j certain force acts on it. The work
done by the force on the particle during
displacement is:
(a) 20 J
(b) 25 J
(c) zero
(d) 18 J
Solution :
 
Displacement d  r2 – r1  i – j
     
Fd  5  i j  i – j   0

 


 

Hence answer is (c)
Class Exercise - 3
A 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
1 

(a) zero
(b)  a  bd  d

(c) 2a + bd
2

(d) (a + 2bd)d
Solution :
Force is variable.
d

bx 2 
bd 2
W   Fdx   (a  bx)dx  ax 
  ad 
2 
2


0
0
0
d
d
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
Solution :
Friction always acts opposite to the
displacement.
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
Solution :
Statement characterizes kinetic energy.
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
f
vt
mg
Sin 
mg

Class Exercise - 7
  
A force F  –K  y i  x j  (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:
(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 




y
Displacement along x = 0
a
2
 W   Fydy  –Ka
0
2
O
1
a
a
x
Class Exercise - 8
Under the action of a force, a 2 kg
body moves such that its position
t3
x as a function of time is given by x 
3
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
t3 dx
d2 x
2
x ,
t , Fm
 2mt
2
3 dt
dt
x
2
2
0
0
0
4
mt
W   Fdx   2mt  t 2dt  2m t3dt 
 16 J
2
Hence answer is (c)
Class Exercise - 9
There 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)
(c)
2Rg
(d)
2Rg
Rg
Solution
PE = mgR
R
KE = 1 mv 2
2
PE = mgR
KE 
1
mv 2
2
1
 mv 2  mgR  v  2Rg
2
Hence answer is (b)
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.
4


 cos37  5 , g  10 m/s 


(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.
Mg sin37° = m N
\ Work done by gravity = Work done
against friction = Mg sin37° x s
cos37 
4
3
, sin37° 
5
5
 W  0.25  10 
3
1
5
= 1.5 J
Hence answer is (c)
N
F=mN
Mg sin37o
Mg cos37o
37o
mg
Thank you