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# DC Pandey Solutions: Work, energy and Power- 2 - Notes | Study DC Pandey Solutions for JEE Physics - JEE

``` Page 1

21. NR mgR mv
s
cos ( ) 180
1
2
2
° - + = a
( cos ) ( cos ) mg R mgR mv
s
a a - + =
1
2
2
or
1
2
1
2 2
mv mgR
s
= - ( cos ) a
or v gR
s
2 2
2 = sin a
or v gR
s
= 2 sin a
(v
s
is the speed with which sphere hits
ground)

1
2
1
2
2 2
mv mgR mv
w s
= -
= - mgR m gR
1
2
2
2
sin a
= - mgR ( sin ) 1
2
a = mgR cos
2
a
\        v gR
w
= 2 cos a
(v
w
is the speed of wedge when the sphere
hits ground)
22. For 45 kg mass to drop 12 mm, the
increase in length of the spring will be
24 mm.
Now, decrease in PE of 45 kg mass
= Increase in KE of 45 kg mass + Increase
in PE of spring
45 12 10
1
2
45
1
2
1050
3 2
´ ´ ´ = ´ ´ + ´
-
9.8 v
´ + - ´
-
[( ) ] 75 24 75 10
2 2 6
i.e., 5.292 22.5 2.192 = + v
2
or 22.5 3.0996 v
2
=
or       v
2
= 0.13776
or            v =
-
0.371 ms
1
(b) With friction (when mechanical
energy does not remain conserved)
23. De crease in PE of 1 kg mass
= Work done against friction due to 4 kg
mass + Increase in KE of 1 kg mass
1 10 1 4 10 2
1
2
1
2
´ ´ = ´ ´ ´ + ´ ´ m
k
( ) 0.3
or 10 80 = + m
k
0.045
or m
k
=
- 10
80
0.045
=0.124
24. f = force of friction while disc in slipping
over inclined surface = ° m mg cos 30
f ¢ = force of friction while disc is slipping
over plane surface = m mg
Now, decreases in PE of disc = Work done
against frictional force
mg
s
fs f
2
= + ¢ ( ) 0.5
or mg
s
mg s mg
2
30 = ° + ( cos ) ( ) m m 0.5
or mgs mg ( cos ) 0.5 (0.5) - ° = m m 30
Þ s =
´
- °
m
m
0.5
0.5 cos 30

=
´
- ´
0.15 0.5
0.5 0.15
3
2
=0.2027 m
Work performed by frictional forces over
the whole distance
= - mg
s
2
= - ´ ´
50
1000
10
2
0.2027
= -0.051 J
25. m g m g T m a
A A A
sin cos q m q - - = 2
or 300
3
5
300
4
5
2 30 ´ - ´ ´ - = 0.2 T a
118 | Mechanics-1
0.50 m
Stops
f'
30°
s (say)
f
N
s/2
a
2R
N
a
R
mg
a
3
mg
B
T
T
2T
T
T
mmg cos q
A
T
T
a
A
S
4
q
m g cos q
A
2T
a
B
Page 2

21. NR mgR mv
s
cos ( ) 180
1
2
2
° - + = a
( cos ) ( cos ) mg R mgR mv
s
a a - + =
1
2
2
or
1
2
1
2 2
mv mgR
s
= - ( cos ) a
or v gR
s
2 2
2 = sin a
or v gR
s
= 2 sin a
(v
s
is the speed with which sphere hits
ground)

1
2
1
2
2 2
mv mgR mv
w s
= -
= - mgR m gR
1
2
2
2
sin a
= - mgR ( sin ) 1
2
a = mgR cos
2
a
\        v gR
w
= 2 cos a
(v
w
is the speed of wedge when the sphere
hits ground)
22. For 45 kg mass to drop 12 mm, the
increase in length of the spring will be
24 mm.
Now, decrease in PE of 45 kg mass
= Increase in KE of 45 kg mass + Increase
in PE of spring
45 12 10
1
2
45
1
2
1050
3 2
´ ´ ´ = ´ ´ + ´
-
9.8 v
´ + - ´
-
[( ) ] 75 24 75 10
2 2 6
i.e., 5.292 22.5 2.192 = + v
2
or 22.5 3.0996 v
2
=
or       v
2
= 0.13776
or            v =
-
0.371 ms
1
(b) With friction (when mechanical
energy does not remain conserved)
23. De crease in PE of 1 kg mass
= Work done against friction due to 4 kg
mass + Increase in KE of 1 kg mass
1 10 1 4 10 2
1
2
1
2
´ ´ = ´ ´ ´ + ´ ´ m
k
( ) 0.3
or 10 80 = + m
k
0.045
or m
k
=
- 10
80
0.045
=0.124
24. f = force of friction while disc in slipping
over inclined surface = ° m mg cos 30
f ¢ = force of friction while disc is slipping
over plane surface = m mg
Now, decreases in PE of disc = Work done
against frictional force
mg
s
fs f
2
= + ¢ ( ) 0.5
or mg
s
mg s mg
2
30 = ° + ( cos ) ( ) m m 0.5
or mgs mg ( cos ) 0.5 (0.5) - ° = m m 30
Þ s =
´
- °
m
m
0.5
0.5 cos 30

=
´
- ´
0.15 0.5
0.5 0.15
3
2
=0.2027 m
Work performed by frictional forces over
the whole distance
= - mg
s
2
= - ´ ´
50
1000
10
2
0.2027
= -0.051 J
25. m g m g T m a
A A A
sin cos q m q - - = 2
or 300
3
5
300
4
5
2 30 ´ - ´ ´ - = 0.2 T a
118 | Mechanics-1
0.50 m
Stops
f'
30°
s (say)
f
N
s/2
a
2R
N
a
R
mg
a
3
mg
B
T
T
2T
T
T
mmg cos q
A
T
T
a
A
S
4
q
m g cos q
A
2T
a
B
or 180 48 2 30 - - = T a
or 132 2 30 - = T a …(i)
Also, T m g m a
B B
- =
or T a - = 50 5
or 2 100 10 T a - = …(ii)
Adding Eq. (i) and Eq. (ii),
32 40 = a
or a =
-
0.8 ms
2
Speed ( ) v of block A after it moves 1 m
down the plane
v as
2
2 =
or v
2
2 1 = ´ ´ 0.8
or v =
-
1.12 ms
1
26. Work done by frictional force acting on
block
= -m mgs
= - ´ ´ ´ 0.25 9.8 7.8 3 5 .
= - 66.88 J
\ Increase in thermal energy of block-floor
system
= 66.88 J
As the block stopped after traversing
7.8 m on rough floor the maximum kinetic
energy of the block would be 66.88 J (just
before entering the rough surface).
Maximum PE of spring
= Maximum KE of block

1
2
2
kx
max
= 66.88
\ x
max
.
=
´ 2 66 88
640
m
Maximum compression in the spring
= 0.457 m
27. Decrease in PE of mass m
2
= Work done
against friction by mass m
1
+ Increase KE
of mass m
1
+ Increase in KE of mass m
2
m g m g m v m v
2 1 1
2
2 2
2
4 4
1
2
1
2
´ = ´ + + m
5 10 4 10 10 4
1
2
10 5
2
´ ´ = ´ ´ ´ + + 0.2 ( )v
Solving, v =
-
4
1
ms
Three types of Equilibrium
28. (a) F
dU
dr
= -
Point
dU
dr
F
A + ive - ive
B + ive - ive
C - ive + ive
D - ive + ive
E zero zero
(b) x = 2 m point is of unstable equilibrium
(U being + ive)
x=6 m point is of stable equilibrium
(U being lowest - ive)
29. U
x
x = - +
3
3
4 6
For U to be maximum (for unstable
equilibrium) and minimum (for stable
equilibrium)
dU
dx
=0
i.e.,
d
dx
x
x
3
3
4 6 0 - +
æ
è
ç
ö
ø
÷ =
or x
2
4 0 - =
or x=±2
dU
dx
d
dx
x x
2
2
4
4 2 = - = ( )
At x=+2 m,
d U
dx
2
2
2 2 4 = ´ + = + ( )
\ U is minium.
At x = - 2 m,
d U
dx
2
2
2 2 4 = ´ - = - ( )
\ U is maximum.
\ x = + 2 m point of stable equilibrium.
x = - 2 m point of unstable equilibrium.
Work, Energy and Power | 119
m = 5 kg
2
m
1
10 kg
Page 3

21. NR mgR mv
s
cos ( ) 180
1
2
2
° - + = a
( cos ) ( cos ) mg R mgR mv
s
a a - + =
1
2
2
or
1
2
1
2 2
mv mgR
s
= - ( cos ) a
or v gR
s
2 2
2 = sin a
or v gR
s
= 2 sin a
(v
s
is the speed with which sphere hits
ground)

1
2
1
2
2 2
mv mgR mv
w s
= -
= - mgR m gR
1
2
2
2
sin a
= - mgR ( sin ) 1
2
a = mgR cos
2
a
\        v gR
w
= 2 cos a
(v
w
is the speed of wedge when the sphere
hits ground)
22. For 45 kg mass to drop 12 mm, the
increase in length of the spring will be
24 mm.
Now, decrease in PE of 45 kg mass
= Increase in KE of 45 kg mass + Increase
in PE of spring
45 12 10
1
2
45
1
2
1050
3 2
´ ´ ´ = ´ ´ + ´
-
9.8 v
´ + - ´
-
[( ) ] 75 24 75 10
2 2 6
i.e., 5.292 22.5 2.192 = + v
2
or 22.5 3.0996 v
2
=
or       v
2
= 0.13776
or            v =
-
0.371 ms
1
(b) With friction (when mechanical
energy does not remain conserved)
23. De crease in PE of 1 kg mass
= Work done against friction due to 4 kg
mass + Increase in KE of 1 kg mass
1 10 1 4 10 2
1
2
1
2
´ ´ = ´ ´ ´ + ´ ´ m
k
( ) 0.3
or 10 80 = + m
k
0.045
or m
k
=
- 10
80
0.045
=0.124
24. f = force of friction while disc in slipping
over inclined surface = ° m mg cos 30
f ¢ = force of friction while disc is slipping
over plane surface = m mg
Now, decreases in PE of disc = Work done
against frictional force
mg
s
fs f
2
= + ¢ ( ) 0.5
or mg
s
mg s mg
2
30 = ° + ( cos ) ( ) m m 0.5
or mgs mg ( cos ) 0.5 (0.5) - ° = m m 30
Þ s =
´
- °
m
m
0.5
0.5 cos 30

=
´
- ´
0.15 0.5
0.5 0.15
3
2
=0.2027 m
Work performed by frictional forces over
the whole distance
= - mg
s
2
= - ´ ´
50
1000
10
2
0.2027
= -0.051 J
25. m g m g T m a
A A A
sin cos q m q - - = 2
or 300
3
5
300
4
5
2 30 ´ - ´ ´ - = 0.2 T a
118 | Mechanics-1
0.50 m
Stops
f'
30°
s (say)
f
N
s/2
a
2R
N
a
R
mg
a
3
mg
B
T
T
2T
T
T
mmg cos q
A
T
T
a
A
S
4
q
m g cos q
A
2T
a
B
or 180 48 2 30 - - = T a
or 132 2 30 - = T a …(i)
Also, T m g m a
B B
- =
or T a - = 50 5
or 2 100 10 T a - = …(ii)
Adding Eq. (i) and Eq. (ii),
32 40 = a
or a =
-
0.8 ms
2
Speed ( ) v of block A after it moves 1 m
down the plane
v as
2
2 =
or v
2
2 1 = ´ ´ 0.8
or v =
-
1.12 ms
1
26. Work done by frictional force acting on
block
= -m mgs
= - ´ ´ ´ 0.25 9.8 7.8 3 5 .
= - 66.88 J
\ Increase in thermal energy of block-floor
system
= 66.88 J
As the block stopped after traversing
7.8 m on rough floor the maximum kinetic
energy of the block would be 66.88 J (just
before entering the rough surface).
Maximum PE of spring
= Maximum KE of block

1
2
2
kx
max
= 66.88
\ x
max
.
=
´ 2 66 88
640
m
Maximum compression in the spring
= 0.457 m
27. Decrease in PE of mass m
2
= Work done
against friction by mass m
1
+ Increase KE
of mass m
1
+ Increase in KE of mass m
2
m g m g m v m v
2 1 1
2
2 2
2
4 4
1
2
1
2
´ = ´ + + m
5 10 4 10 10 4
1
2
10 5
2
´ ´ = ´ ´ ´ + + 0.2 ( )v
Solving, v =
-
4
1
ms
Three types of Equilibrium
28. (a) F
dU
dr
= -
Point
dU
dr
F
A + ive - ive
B + ive - ive
C - ive + ive
D - ive + ive
E zero zero
(b) x = 2 m point is of unstable equilibrium
(U being + ive)
x=6 m point is of stable equilibrium
(U being lowest - ive)
29. U
x
x = - +
3
3
4 6
For U to be maximum (for unstable
equilibrium) and minimum (for stable
equilibrium)
dU
dx
=0
i.e.,
d
dx
x
x
3
3
4 6 0 - +
æ
è
ç
ö
ø
÷ =
or x
2
4 0 - =
or x=±2
dU
dx
d
dx
x x
2
2
4
4 2 = - = ( )
At x=+2 m,
d U
dx
2
2
2 2 4 = ´ + = + ( )
\ U is minium.
At x = - 2 m,
d U
dx
2
2
2 2 4 = ´ - = - ( )
\ U is maximum.
\ x = + 2 m point of stable equilibrium.
x = - 2 m point of unstable equilibrium.
Work, Energy and Power | 119
m = 5 kg
2
m
1
10 kg
30. F
dU
dx
= -
i.e., U F dx = -
ò
= - (Area under F x - graph)
The corresponding U Vs x graph will be as
shown in figure
Thus, point C corresponds to stable
equilibrium and points A and E
correspond to unstable equilibrium.
31. - q charge placed at origin is in
equilibrium as is two equal and opposite
forces act on it.
(a)But, if we displace it slightly say
towards +ive x side, the force on it due
to charge to B will decreases while that
due to A will increase.
Due to net force on - q towards right the
change - q will never come back to original
O, its origin position.
Thus, the equilibrium of the charge -q
is unstable if it is slightly displaced along
x-axis.
(b) If charge - q is displaced slightly along
Y axis, the net force on it will be along
its original position. And as such the
equilibrium of the - q is stable.
32. (a) Velocity at t = 0 s is 0 ms
-1
Velocity at t=2 s is 8 ms
-1
(using
v at = + 0 )
\ v
av
ms =
-
4
1
(as acceleration is constant)
P F v
av av
= ´
=mav
av

= ´ ´ = 1 4 4 16 W
(b) Velocity at t = 4 s is 16 ms
-1
(using
v u at = + )
\ Instantaneous power of the net force at
t = 4 s will be
P mav =
= ´ ´ 1 4 16
=64 W
33. Power = F v
min max
P rv =
max
\ v
P
r
max
=
Objective Questions (Level 1)
Single Correct Option
1. In KE =
1
2
2
mv
m is always +ive and v
2
is +ive.
(even if v is - ive).
\ KE is always + ive.
Below reference level ( ) PE = 0 PE is - ive.
\ Mechanical energy which is sum of KE
and PE may be - ive.
\ Correct option is (a).
120 | Mechanics-1
F
x
x
E C
D
A
B
U
+
–
+ q
– q
F A
(+ a, 0, 0)
F
x
+ q
B
(– a, 0, 0)
y
+ q
– q
F
net
A
F'
+ q
B
(– a, 0, 0)
F'
O
+ q – q
F
2
F > F
12
O
r
F
max
Initially
r
F = (r)
min
Finally
v
max
Page 4

21. NR mgR mv
s
cos ( ) 180
1
2
2
° - + = a
( cos ) ( cos ) mg R mgR mv
s
a a - + =
1
2
2
or
1
2
1
2 2
mv mgR
s
= - ( cos ) a
or v gR
s
2 2
2 = sin a
or v gR
s
= 2 sin a
(v
s
is the speed with which sphere hits
ground)

1
2
1
2
2 2
mv mgR mv
w s
= -
= - mgR m gR
1
2
2
2
sin a
= - mgR ( sin ) 1
2
a = mgR cos
2
a
\        v gR
w
= 2 cos a
(v
w
is the speed of wedge when the sphere
hits ground)
22. For 45 kg mass to drop 12 mm, the
increase in length of the spring will be
24 mm.
Now, decrease in PE of 45 kg mass
= Increase in KE of 45 kg mass + Increase
in PE of spring
45 12 10
1
2
45
1
2
1050
3 2
´ ´ ´ = ´ ´ + ´
-
9.8 v
´ + - ´
-
[( ) ] 75 24 75 10
2 2 6
i.e., 5.292 22.5 2.192 = + v
2
or 22.5 3.0996 v
2
=
or       v
2
= 0.13776
or            v =
-
0.371 ms
1
(b) With friction (when mechanical
energy does not remain conserved)
23. De crease in PE of 1 kg mass
= Work done against friction due to 4 kg
mass + Increase in KE of 1 kg mass
1 10 1 4 10 2
1
2
1
2
´ ´ = ´ ´ ´ + ´ ´ m
k
( ) 0.3
or 10 80 = + m
k
0.045
or m
k
=
- 10
80
0.045
=0.124
24. f = force of friction while disc in slipping
over inclined surface = ° m mg cos 30
f ¢ = force of friction while disc is slipping
over plane surface = m mg
Now, decreases in PE of disc = Work done
against frictional force
mg
s
fs f
2
= + ¢ ( ) 0.5
or mg
s
mg s mg
2
30 = ° + ( cos ) ( ) m m 0.5
or mgs mg ( cos ) 0.5 (0.5) - ° = m m 30
Þ s =
´
- °
m
m
0.5
0.5 cos 30

=
´
- ´
0.15 0.5
0.5 0.15
3
2
=0.2027 m
Work performed by frictional forces over
the whole distance
= - mg
s
2
= - ´ ´
50
1000
10
2
0.2027
= -0.051 J
25. m g m g T m a
A A A
sin cos q m q - - = 2
or 300
3
5
300
4
5
2 30 ´ - ´ ´ - = 0.2 T a
118 | Mechanics-1
0.50 m
Stops
f'
30°
s (say)
f
N
s/2
a
2R
N
a
R
mg
a
3
mg
B
T
T
2T
T
T
mmg cos q
A
T
T
a
A
S
4
q
m g cos q
A
2T
a
B
or 180 48 2 30 - - = T a
or 132 2 30 - = T a …(i)
Also, T m g m a
B B
- =
or T a - = 50 5
or 2 100 10 T a - = …(ii)
Adding Eq. (i) and Eq. (ii),
32 40 = a
or a =
-
0.8 ms
2
Speed ( ) v of block A after it moves 1 m
down the plane
v as
2
2 =
or v
2
2 1 = ´ ´ 0.8
or v =
-
1.12 ms
1
26. Work done by frictional force acting on
block
= -m mgs
= - ´ ´ ´ 0.25 9.8 7.8 3 5 .
= - 66.88 J
\ Increase in thermal energy of block-floor
system
= 66.88 J
As the block stopped after traversing
7.8 m on rough floor the maximum kinetic
energy of the block would be 66.88 J (just
before entering the rough surface).
Maximum PE of spring
= Maximum KE of block

1
2
2
kx
max
= 66.88
\ x
max
.
=
´ 2 66 88
640
m
Maximum compression in the spring
= 0.457 m
27. Decrease in PE of mass m
2
= Work done
against friction by mass m
1
+ Increase KE
of mass m
1
+ Increase in KE of mass m
2
m g m g m v m v
2 1 1
2
2 2
2
4 4
1
2
1
2
´ = ´ + + m
5 10 4 10 10 4
1
2
10 5
2
´ ´ = ´ ´ ´ + + 0.2 ( )v
Solving, v =
-
4
1
ms
Three types of Equilibrium
28. (a) F
dU
dr
= -
Point
dU
dr
F
A + ive - ive
B + ive - ive
C - ive + ive
D - ive + ive
E zero zero
(b) x = 2 m point is of unstable equilibrium
(U being + ive)
x=6 m point is of stable equilibrium
(U being lowest - ive)
29. U
x
x = - +
3
3
4 6
For U to be maximum (for unstable
equilibrium) and minimum (for stable
equilibrium)
dU
dx
=0
i.e.,
d
dx
x
x
3
3
4 6 0 - +
æ
è
ç
ö
ø
÷ =
or x
2
4 0 - =
or x=±2
dU
dx
d
dx
x x
2
2
4
4 2 = - = ( )
At x=+2 m,
d U
dx
2
2
2 2 4 = ´ + = + ( )
\ U is minium.
At x = - 2 m,
d U
dx
2
2
2 2 4 = ´ - = - ( )
\ U is maximum.
\ x = + 2 m point of stable equilibrium.
x = - 2 m point of unstable equilibrium.
Work, Energy and Power | 119
m = 5 kg
2
m
1
10 kg
30. F
dU
dx
= -
i.e., U F dx = -
ò
= - (Area under F x - graph)
The corresponding U Vs x graph will be as
shown in figure
Thus, point C corresponds to stable
equilibrium and points A and E
correspond to unstable equilibrium.
31. - q charge placed at origin is in
equilibrium as is two equal and opposite
forces act on it.
(a)But, if we displace it slightly say
towards +ive x side, the force on it due
to charge to B will decreases while that
due to A will increase.
Due to net force on - q towards right the
change - q will never come back to original
O, its origin position.
Thus, the equilibrium of the charge -q
is unstable if it is slightly displaced along
x-axis.
(b) If charge - q is displaced slightly along
Y axis, the net force on it will be along
its original position. And as such the
equilibrium of the - q is stable.
32. (a) Velocity at t = 0 s is 0 ms
-1
Velocity at t=2 s is 8 ms
-1
(using
v at = + 0 )
\ v
av
ms =
-
4
1
(as acceleration is constant)
P F v
av av
= ´
=mav
av

= ´ ´ = 1 4 4 16 W
(b) Velocity at t = 4 s is 16 ms
-1
(using
v u at = + )
\ Instantaneous power of the net force at
t = 4 s will be
P mav =
= ´ ´ 1 4 16
=64 W
33. Power = F v
min max
P rv =
max
\ v
P
r
max
=
Objective Questions (Level 1)
Single Correct Option
1. In KE =
1
2
2
mv
m is always +ive and v
2
is +ive.
(even if v is - ive).
\ KE is always + ive.
Below reference level ( ) PE = 0 PE is - ive.
\ Mechanical energy which is sum of KE
and PE may be - ive.
\ Correct option is (a).
120 | Mechanics-1
F
x
x
E C
D
A
B
U
+
–
+ q
– q
F A
(+ a, 0, 0)
F
x
+ q
B
(– a, 0, 0)
y
+ q
– q
F
net
A
F'
+ q
B
(– a, 0, 0)
F'
O
+ q – q
F
2
F > F
12
O
r
F
max
Initially
r
F = (r)
min
Finally
v
max
2. Yes, this is work-en ergy the o rem.
\ Correct option is (a).
3. On a body placed on a rough surface if an
external force is applied the static body
does not move then the work done by
frictional force will be zero.
4. W = ×
® ®
F r
21
= × -
® ® ®
F r r ( )
2 1
= + + × - + - + + ( ) [( ) ( )]
^ ^ ^ ^ ^ ^ ^ ^ ^
i j k i j k i j k 2 3 2
= + + × - + ( ) ( )
^ ^ ^ ^ ^
i j k j k 2 3 2
= - + 4 3
= - 1 J
\ Correct option is (b).
5. \ W F dx =
ò
= - +
ò
7 2 3
0
5
2
x x dx
= - + = [ ] 7 135
2 3
0
5
x x x J
\ Correct option is (d).
6. P = ×
® ®
F v
= + + × - + ( ) ( )
^ ^ ^ ^ ^ ^
10 10 20 5 3 6 i j k i j k
= - + 50 30 120
=140 W
\ Correct option is (c).
7. Work done in displacing the body
= Area under the curve
= ´ + ´ + ´ - +
´
æ
è
ç
ö
ø
÷ ( ) ( ) ( ) 1 10 1 5 1 5
1 10
2
= 15 J
\ Correct option is (b).
8. P
W
t
mgh mv
t
= =
+
1
2
2
=
´ ´ +
´ ´
-
-
800 10 10
1
2
800 20
1
2
1 2
kg ms m
kg ms
min
( )
=
´ ´ + ´ ( ) [ ( ) ] 800 10 10 400 20
60
2
=
+ 80000 160000
60
= 4000 W
Option (c) is correct.
9. x
t
=
3
3
\ v
dx
dt
t = =
2
and a
dv
dt
t = = 2
F ma t t = = ´ = 2 2 4
W F v = = ´ = 4 4
2 3
t t t
\ Work done by force in first two seconds
= ×
=
=
ò
4
3
0
2
t dt
t
t
= ×
é
ë
ê
ù
û
ú
4
4
4
0
2
t

=16 J
\ Correct option is (c).
10. Range = 4 ´ height
or
u
g
u
g
2 2 2
2
4
2
sin sin q q
= ×
or sin sin 2 2
2
q q =
or 2 2 0
2
sin cos sin q q q - =
or sin (cos sin ) q q q - = 0
or cos sin q q - = 0 (as sin q ¹ 0)
or tan q = 1
i.e., q = ° 45
Now, K = KE at highest point
=
1
2
2
m u ( cos ) q
=
1
2
2 2
mu cos q
= °
1
2
45
2 2
mu cos
= ×
1
2
1
2
2
mu
\
1
2
2
2
mu K =
i.e., Initial KE = 2K
\ Correct option is (b).
Work, Energy and Power | 121
Page 5

21. NR mgR mv
s
cos ( ) 180
1
2
2
° - + = a
( cos ) ( cos ) mg R mgR mv
s
a a - + =
1
2
2
or
1
2
1
2 2
mv mgR
s
= - ( cos ) a
or v gR
s
2 2
2 = sin a
or v gR
s
= 2 sin a
(v
s
is the speed with which sphere hits
ground)

1
2
1
2
2 2
mv mgR mv
w s
= -
= - mgR m gR
1
2
2
2
sin a
= - mgR ( sin ) 1
2
a = mgR cos
2
a
\        v gR
w
= 2 cos a
(v
w
is the speed of wedge when the sphere
hits ground)
22. For 45 kg mass to drop 12 mm, the
increase in length of the spring will be
24 mm.
Now, decrease in PE of 45 kg mass
= Increase in KE of 45 kg mass + Increase
in PE of spring
45 12 10
1
2
45
1
2
1050
3 2
´ ´ ´ = ´ ´ + ´
-
9.8 v
´ + - ´
-
[( ) ] 75 24 75 10
2 2 6
i.e., 5.292 22.5 2.192 = + v
2
or 22.5 3.0996 v
2
=
or       v
2
= 0.13776
or            v =
-
0.371 ms
1
(b) With friction (when mechanical
energy does not remain conserved)
23. De crease in PE of 1 kg mass
= Work done against friction due to 4 kg
mass + Increase in KE of 1 kg mass
1 10 1 4 10 2
1
2
1
2
´ ´ = ´ ´ ´ + ´ ´ m
k
( ) 0.3
or 10 80 = + m
k
0.045
or m
k
=
- 10
80
0.045
=0.124
24. f = force of friction while disc in slipping
over inclined surface = ° m mg cos 30
f ¢ = force of friction while disc is slipping
over plane surface = m mg
Now, decreases in PE of disc = Work done
against frictional force
mg
s
fs f
2
= + ¢ ( ) 0.5
or mg
s
mg s mg
2
30 = ° + ( cos ) ( ) m m 0.5
or mgs mg ( cos ) 0.5 (0.5) - ° = m m 30
Þ s =
´
- °
m
m
0.5
0.5 cos 30

=
´
- ´
0.15 0.5
0.5 0.15
3
2
=0.2027 m
Work performed by frictional forces over
the whole distance
= - mg
s
2
= - ´ ´
50
1000
10
2
0.2027
= -0.051 J
25. m g m g T m a
A A A
sin cos q m q - - = 2
or 300
3
5
300
4
5
2 30 ´ - ´ ´ - = 0.2 T a
118 | Mechanics-1
0.50 m
Stops
f'
30°
s (say)
f
N
s/2
a
2R
N
a
R
mg
a
3
mg
B
T
T
2T
T
T
mmg cos q
A
T
T
a
A
S
4
q
m g cos q
A
2T
a
B
or 180 48 2 30 - - = T a
or 132 2 30 - = T a …(i)
Also, T m g m a
B B
- =
or T a - = 50 5
or 2 100 10 T a - = …(ii)
Adding Eq. (i) and Eq. (ii),
32 40 = a
or a =
-
0.8 ms
2
Speed ( ) v of block A after it moves 1 m
down the plane
v as
2
2 =
or v
2
2 1 = ´ ´ 0.8
or v =
-
1.12 ms
1
26. Work done by frictional force acting on
block
= -m mgs
= - ´ ´ ´ 0.25 9.8 7.8 3 5 .
= - 66.88 J
\ Increase in thermal energy of block-floor
system
= 66.88 J
As the block stopped after traversing
7.8 m on rough floor the maximum kinetic
energy of the block would be 66.88 J (just
before entering the rough surface).
Maximum PE of spring
= Maximum KE of block

1
2
2
kx
max
= 66.88
\ x
max
.
=
´ 2 66 88
640
m
Maximum compression in the spring
= 0.457 m
27. Decrease in PE of mass m
2
= Work done
against friction by mass m
1
+ Increase KE
of mass m
1
+ Increase in KE of mass m
2
m g m g m v m v
2 1 1
2
2 2
2
4 4
1
2
1
2
´ = ´ + + m
5 10 4 10 10 4
1
2
10 5
2
´ ´ = ´ ´ ´ + + 0.2 ( )v
Solving, v =
-
4
1
ms
Three types of Equilibrium
28. (a) F
dU
dr
= -
Point
dU
dr
F
A + ive - ive
B + ive - ive
C - ive + ive
D - ive + ive
E zero zero
(b) x = 2 m point is of unstable equilibrium
(U being + ive)
x=6 m point is of stable equilibrium
(U being lowest - ive)
29. U
x
x = - +
3
3
4 6
For U to be maximum (for unstable
equilibrium) and minimum (for stable
equilibrium)
dU
dx
=0
i.e.,
d
dx
x
x
3
3
4 6 0 - +
æ
è
ç
ö
ø
÷ =
or x
2
4 0 - =
or x=±2
dU
dx
d
dx
x x
2
2
4
4 2 = - = ( )
At x=+2 m,
d U
dx
2
2
2 2 4 = ´ + = + ( )
\ U is minium.
At x = - 2 m,
d U
dx
2
2
2 2 4 = ´ - = - ( )
\ U is maximum.
\ x = + 2 m point of stable equilibrium.
x = - 2 m point of unstable equilibrium.
Work, Energy and Power | 119
m = 5 kg
2
m
1
10 kg
30. F
dU
dx
= -
i.e., U F dx = -
ò
= - (Area under F x - graph)
The corresponding U Vs x graph will be as
shown in figure
Thus, point C corresponds to stable
equilibrium and points A and E
correspond to unstable equilibrium.
31. - q charge placed at origin is in
equilibrium as is two equal and opposite
forces act on it.
(a)But, if we displace it slightly say
towards +ive x side, the force on it due
to charge to B will decreases while that
due to A will increase.
Due to net force on - q towards right the
change - q will never come back to original
O, its origin position.
Thus, the equilibrium of the charge -q
is unstable if it is slightly displaced along
x-axis.
(b) If charge - q is displaced slightly along
Y axis, the net force on it will be along
its original position. And as such the
equilibrium of the - q is stable.
32. (a) Velocity at t = 0 s is 0 ms
-1
Velocity at t=2 s is 8 ms
-1
(using
v at = + 0 )
\ v
av
ms =
-
4
1
(as acceleration is constant)
P F v
av av
= ´
=mav
av

= ´ ´ = 1 4 4 16 W
(b) Velocity at t = 4 s is 16 ms
-1
(using
v u at = + )
\ Instantaneous power of the net force at
t = 4 s will be
P mav =
= ´ ´ 1 4 16
=64 W
33. Power = F v
min max
P rv =
max
\ v
P
r
max
=
Objective Questions (Level 1)
Single Correct Option
1. In KE =
1
2
2
mv
m is always +ive and v
2
is +ive.
(even if v is - ive).
\ KE is always + ive.
Below reference level ( ) PE = 0 PE is - ive.
\ Mechanical energy which is sum of KE
and PE may be - ive.
\ Correct option is (a).
120 | Mechanics-1
F
x
x
E C
D
A
B
U
+
–
+ q
– q
F A
(+ a, 0, 0)
F
x
+ q
B
(– a, 0, 0)
y
+ q
– q
F
net
A
F'
+ q
B
(– a, 0, 0)
F'
O
+ q – q
F
2
F > F
12
O
r
F
max
Initially
r
F = (r)
min
Finally
v
max
2. Yes, this is work-en ergy the o rem.
\ Correct option is (a).
3. On a body placed on a rough surface if an
external force is applied the static body
does not move then the work done by
frictional force will be zero.
4. W = ×
® ®
F r
21
= × -
® ® ®
F r r ( )
2 1
= + + × - + - + + ( ) [( ) ( )]
^ ^ ^ ^ ^ ^ ^ ^ ^
i j k i j k i j k 2 3 2
= + + × - + ( ) ( )
^ ^ ^ ^ ^
i j k j k 2 3 2
= - + 4 3
= - 1 J
\ Correct option is (b).
5. \ W F dx =
ò
= - +
ò
7 2 3
0
5
2
x x dx
= - + = [ ] 7 135
2 3
0
5
x x x J
\ Correct option is (d).
6. P = ×
® ®
F v
= + + × - + ( ) ( )
^ ^ ^ ^ ^ ^
10 10 20 5 3 6 i j k i j k
= - + 50 30 120
=140 W
\ Correct option is (c).
7. Work done in displacing the body
= Area under the curve
= ´ + ´ + ´ - +
´
æ
è
ç
ö
ø
÷ ( ) ( ) ( ) 1 10 1 5 1 5
1 10
2
= 15 J
\ Correct option is (b).
8. P
W
t
mgh mv
t
= =
+
1
2
2
=
´ ´ +
´ ´
-
-
800 10 10
1
2
800 20
1
2
1 2
kg ms m
kg ms
min
( )
=
´ ´ + ´ ( ) [ ( ) ] 800 10 10 400 20
60
2
=
+ 80000 160000
60
= 4000 W
Option (c) is correct.
9. x
t
=
3
3
\ v
dx
dt
t = =
2
and a
dv
dt
t = = 2
F ma t t = = ´ = 2 2 4
W F v = = ´ = 4 4
2 3
t t t
\ Work done by force in first two seconds
= ×
=
=
ò
4
3
0
2
t dt
t
t
= ×
é
ë
ê
ù
û
ú
4
4
4
0
2
t

=16 J
\ Correct option is (c).
10. Range = 4 ´ height
or
u
g
u
g
2 2 2
2
4
2
sin sin q q
= ×
or sin sin 2 2
2
q q =
or 2 2 0
2
sin cos sin q q q - =
or sin (cos sin ) q q q - = 0
or cos sin q q - = 0 (as sin q ¹ 0)
or tan q = 1
i.e., q = ° 45
Now, K = KE at highest point
=
1
2
2
m u ( cos ) q
=
1
2
2 2
mu cos q
= °
1
2
45
2 2
mu cos
= ×
1
2
1
2
2
mu
\
1
2
2
2
mu K =
i.e., Initial KE = 2K
\ Correct option is (b).
Work, Energy and Power | 121
11. P t t = - + 3 2 1
2
i.e.,
dW
dt
t t = - + 3 2 1
2
dW t t dt = - + ( ) 3 2 1
2
W t t dt = - +
ò
( ) 3 2 1
2
2
4
= × - +
é
ë
ê
ù
û
ú
3
3
2
2
3 2
2
4
t t
t
or      = - + [ ] t t t
3 2
2
4
= - + - - + ( ) ( ) 4 4 4 2 2 2
3 2 3 2
= - 52 6
= 46
DK = 46 J
Correct option is (b).
12. K
i
= ´ ´ =
1
2
10 10 500
2
J
Work done by retarding force,
W xdx = -
ò
0.1
20
30
= -
é
ë
ê
ù
û
ú
0.1
x
2
20
30
2
= - ´ - 0.05 [( ) ( ) ] 30 20
2 2
= -25 J
Final kinetic energy = + K W
i
= + - 500 25 ( )
= 475 J
\ Correct option is (a).
13. KE of 12 kg mass : KE of 6 kg mass
=
1
2
1
2
12
2
6
2
m v m v :
[Acceleration being same (equal to g) both
will have same velocities]
= m m
12 6
:
=12 6 :
=2 1 :
14. W k x x = -
1
2
2
2
1
2
[ ]
= ´ ´
æ
è
ç
ö
ø
÷ -
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
1
2
5 10
10
100
5
100
3
2 2
=
´
´
- =
5 10
2 10
100 25
3
4
( ) 18.75 Nm
\ Correct option is (c).
15. x = 2.0 m to 3.5 m
dU
dx
= - = -
6
4
1.5
\ F = + 4 N
x = 3.5 m to 4.5 m

dU
dx
= =
2
1
2
\           F = -2 N
x = 4.5  m to 5.0 m

dU
dx
=0
\          F =0 N
Work done = ´ + - ´ + ( ) ( ) 4 2 1 0 1.5
= 4 J
\
1
2
4
2
mv =
or
1
2
1 4
2
´ ´ = v
\               v =
-
2 2
1
ms
16. KE at highest point
= °
1
2
45
2
m u ( cos )
=
æ
è
ç
ö
ø
÷
1
2
1
2
2
2
mu
=
æ
è
ç
ö
ø
÷
1
2
1
2
2
mu
=
E
2
\ Correct option is (a).
17. Work done by person
= - [Work done by gravitational pull on rope +
gravitational pull on bucket]
= -
æ
è
ç
ö
ø
÷
+ -
é
ë
ê
ù
û
ú
mg
h
Mgh
2
( )
= +
æ
è
ç
ö
ø
÷ M
m
gh
2
\ Correct option is (a).
18.
1
2
2
mv Fx =                  …(i)
1
2
2
mv Fx ¢ = ¢ …(ii)
\
x
x
v
v
¢
=
¢
2
2
=
( ) 2
2
2
v
v
= 4 Þ  x x ¢ = 4
\ Correct option is (b).
122 | Mechanics-1
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## DC Pandey Solutions for JEE Physics

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