Page 1
12. Decrease in KE = 
1
2
0
2
mv
=
1
2
2
mv
Change in mechanical energy
PE + KE =
1
2
2
mv
Decrease in ME is used up in doing work
against friction.
In this case the mechanical energy is
being used up in doing work against
friction and in increasing the PE of the
block.
\ Change ME = 
1
2
2
mv mgh
Thus, assertion is true.
As explained above the reason is false.
(m does not change with the increase in
angle of inclination)
\ Correct option is (c).
Objective Questions (Level 2)
Single Correct Option
1. Increase in KE of bead = Work done by
gravity + work done by force F
\
1
2
2
mv mgh FR = +
(Displacement of force F is R)
= ´ ´ + ´
1
2
10 5 5 5 = 50
\ u
m
=
100
= 200 =

14.14 ms
1
\ Correct option is (a).
2. P Fv =
= mav
=
æ
è
ç
ö
ø
÷
m v
dv
dx
v a
dv
dt
dv
ds
ds
dt
v
dv
ds
= = × =
æ
è
ç
ö
ø
÷
\ ds
m
p
v dv =
2
i.e., ds
m
p
v dv
v
v
ò ò
=
2
2
or s
m
p
v
v
v
=
é
ë
ê
ù
û
ú
3
2
3
= 
m
p
v v
3
2
3 3
[( ) ( ) ] =
7
3
3
mv
p
\ Correct option is (a).
3. Loss of PE of block = Gain in PE of spring
mg d k ( ) sin + ° = ´ 2 30
1
2
2
2
or 10 10 2
1
2
1
2
100 2
2
´ ´ + ´ = ´ ´ ( ) d
d = 2 m
Let, v
A
= velocity of block when it just
touches spring
\
1
2
30
2
mv mgd
A
= ° sin
v
A
2
10 2 2
1
2
= ´ ´ ´
v
A
=

20
1
ms
Correct option is (a).
128  Mechanics1
30°
f'
v'
Stops
PE = mgh
h
v
Slope
CaseI Rough
f
d
A
k
2m
A
30°
d
m
Page 2
12. Decrease in KE = 
1
2
0
2
mv
=
1
2
2
mv
Change in mechanical energy
PE + KE =
1
2
2
mv
Decrease in ME is used up in doing work
against friction.
In this case the mechanical energy is
being used up in doing work against
friction and in increasing the PE of the
block.
\ Change ME = 
1
2
2
mv mgh
Thus, assertion is true.
As explained above the reason is false.
(m does not change with the increase in
angle of inclination)
\ Correct option is (c).
Objective Questions (Level 2)
Single Correct Option
1. Increase in KE of bead = Work done by
gravity + work done by force F
\
1
2
2
mv mgh FR = +
(Displacement of force F is R)
= ´ ´ + ´
1
2
10 5 5 5 = 50
\ u
m
=
100
= 200 =

14.14 ms
1
\ Correct option is (a).
2. P Fv =
= mav
=
æ
è
ç
ö
ø
÷
m v
dv
dx
v a
dv
dt
dv
ds
ds
dt
v
dv
ds
= = × =
æ
è
ç
ö
ø
÷
\ ds
m
p
v dv =
2
i.e., ds
m
p
v dv
v
v
ò ò
=
2
2
or s
m
p
v
v
v
=
é
ë
ê
ù
û
ú
3
2
3
= 
m
p
v v
3
2
3 3
[( ) ( ) ] =
7
3
3
mv
p
\ Correct option is (a).
3. Loss of PE of block = Gain in PE of spring
mg d k ( ) sin + ° = ´ 2 30
1
2
2
2
or 10 10 2
1
2
1
2
100 2
2
´ ´ + ´ = ´ ´ ( ) d
d = 2 m
Let, v
A
= velocity of block when it just
touches spring
\
1
2
30
2
mv mgd
A
= ° sin
v
A
2
10 2 2
1
2
= ´ ´ ´
v
A
=

20
1
ms
Correct option is (a).
128  Mechanics1
30°
f'
v'
Stops
PE = mgh
h
v
Slope
CaseI Rough
f
d
A
k
2m
A
30°
d
m
Work, Energy and Power  129
4. Mass per unit length of chain =
m
R p / 2
dm Rd
m
R
= q
p /2
y
ydm
dm
CM
=
ò
ò
=

ò
R
m
R
Rd
m
0
2
1
2
p
q
p
q
/
( cos )
/
= 
ò
2
1
0
2 R
d
p
q q
p
( cos )
/
= 
é
ë
ê
ù
û
ú
2
2
1
R
p
p
= 
é
ë
ê
ù
û
ú
R 1
2
p
Now,
1
2
2
mv mg y +
CM
or v g y
2
2 =
COM
v gR = 
æ
è
ç
ö
ø
÷ 2 1
2
p
\ Correct option is (c).
5. The moment string is cut
Net force on A mg = 2 (downward)
\ a g
1
2 =
Net force on B = 0
\ a
2
0 =
\ Correct option is (b).
6. Let x be the expansion in the spring.
Increase in PE of spring = Decrease in PE
of block C
1
2
2
1
kx M gx =
i.e., kx M g = 2
1
For block A to remain at rest
kx Mg = m
min
or 2
1
M g Mg = m
min
\ m
min
=
2
1
M
M
\ Correct option is (c).
7. T mg
i
=
KX
mg
mg
i
= =
2
2
When one spring is cut. It means KX
i
becomes zero. Downward acceleration,
a
KX
m
mg
m
g
i
= = =
2 2 2
Now drawing FBD of lower mass :
mg T m a
mg
f
 = = .
2
\ T
mg
f
=
2
or DT T T
mg
f i
=  =
2
8. a
mg mg
m
=
 sin cos q m q
=  g g sin cos q m q
For v to be maximum
dv
dx
= 0
or v
dv
dx
= 0
F (= mg)
mg
mg
A
B
F = mg
T = mg
T
F = kx = mg
mg (weight)
F
mg
Mg
1
A B
kx kx
T T
C
T
T
R (1 – cos q)
CM
y
( )
R 1 –
2
p
q
q
a
x
mmg cos q
mg sin q
Page 3
12. Decrease in KE = 
1
2
0
2
mv
=
1
2
2
mv
Change in mechanical energy
PE + KE =
1
2
2
mv
Decrease in ME is used up in doing work
against friction.
In this case the mechanical energy is
being used up in doing work against
friction and in increasing the PE of the
block.
\ Change ME = 
1
2
2
mv mgh
Thus, assertion is true.
As explained above the reason is false.
(m does not change with the increase in
angle of inclination)
\ Correct option is (c).
Objective Questions (Level 2)
Single Correct Option
1. Increase in KE of bead = Work done by
gravity + work done by force F
\
1
2
2
mv mgh FR = +
(Displacement of force F is R)
= ´ ´ + ´
1
2
10 5 5 5 = 50
\ u
m
=
100
= 200 =

14.14 ms
1
\ Correct option is (a).
2. P Fv =
= mav
=
æ
è
ç
ö
ø
÷
m v
dv
dx
v a
dv
dt
dv
ds
ds
dt
v
dv
ds
= = × =
æ
è
ç
ö
ø
÷
\ ds
m
p
v dv =
2
i.e., ds
m
p
v dv
v
v
ò ò
=
2
2
or s
m
p
v
v
v
=
é
ë
ê
ù
û
ú
3
2
3
= 
m
p
v v
3
2
3 3
[( ) ( ) ] =
7
3
3
mv
p
\ Correct option is (a).
3. Loss of PE of block = Gain in PE of spring
mg d k ( ) sin + ° = ´ 2 30
1
2
2
2
or 10 10 2
1
2
1
2
100 2
2
´ ´ + ´ = ´ ´ ( ) d
d = 2 m
Let, v
A
= velocity of block when it just
touches spring
\
1
2
30
2
mv mgd
A
= ° sin
v
A
2
10 2 2
1
2
= ´ ´ ´
v
A
=

20
1
ms
Correct option is (a).
128  Mechanics1
30°
f'
v'
Stops
PE = mgh
h
v
Slope
CaseI Rough
f
d
A
k
2m
A
30°
d
m
Work, Energy and Power  129
4. Mass per unit length of chain =
m
R p / 2
dm Rd
m
R
= q
p /2
y
ydm
dm
CM
=
ò
ò
=

ò
R
m
R
Rd
m
0
2
1
2
p
q
p
q
/
( cos )
/
= 
ò
2
1
0
2 R
d
p
q q
p
( cos )
/
= 
é
ë
ê
ù
û
ú
2
2
1
R
p
p
= 
é
ë
ê
ù
û
ú
R 1
2
p
Now,
1
2
2
mv mg y +
CM
or v g y
2
2 =
COM
v gR = 
æ
è
ç
ö
ø
÷ 2 1
2
p
\ Correct option is (c).
5. The moment string is cut
Net force on A mg = 2 (downward)
\ a g
1
2 =
Net force on B = 0
\ a
2
0 =
\ Correct option is (b).
6. Let x be the expansion in the spring.
Increase in PE of spring = Decrease in PE
of block C
1
2
2
1
kx M gx =
i.e., kx M g = 2
1
For block A to remain at rest
kx Mg = m
min
or 2
1
M g Mg = m
min
\ m
min
=
2
1
M
M
\ Correct option is (c).
7. T mg
i
=
KX
mg
mg
i
= =
2
2
When one spring is cut. It means KX
i
becomes zero. Downward acceleration,
a
KX
m
mg
m
g
i
= = =
2 2 2
Now drawing FBD of lower mass :
mg T m a
mg
f
 = = .
2
\ T
mg
f
=
2
or DT T T
mg
f i
=  =
2
8. a
mg mg
m
=
 sin cos q m q
=  g g sin cos q m q
For v to be maximum
dv
dx
= 0
or v
dv
dx
= 0
F (= mg)
mg
mg
A
B
F = mg
T = mg
T
F = kx = mg
mg (weight)
F
mg
Mg
1
A B
kx kx
T T
C
T
T
R (1 – cos q)
CM
y
( )
R 1 –
2
p
q
q
a
x
mmg cos q
mg sin q
or
dx
dt
dv
dx
× = 0
or
dv
dt
= 0
or a = 0
i.e., g g sin cos q m q =
or sin cos q m q =
or
3
5
3
10
4
5
= × x
Þ x = =
10
4
2.5 m
\ Correct option is (d).
9. (a) Between points E and F,
dU
dr
is  ive.
Now, F
dU
dr
=  , the force between E and F
will be + ive i.e., repulsive.
(b) At point C the potential energy is
minimum. Thus, C is point of stable
equilibrium.
\ Correct option is (c).
10. Power = ×
® ®
F v
= × +
® ® ®
m t g u g ( )
= ° + + ×
® ®
mg u m t cos ( ) 90 q g g
i.e., P mg u mg t =  + sin q
2
[   ,  u g
® ®
= = u g]
Therefore, the graph between P and t will
be as shown in option (c).
11.
PE of spring due to its compression by x
=  Work done by frictional force when
displaced by 2x
i.e.,
1
2
2
2
kx mg x = m
Þ x
mg
x
=
4 m
\ Correct option is (c).
12. Power delivered by man = ×
® ®
T v
= T vcos q
  T
®
= T
and   v
®
= T
13. f = + 3 4 x y
\ F
x
x
= 
¶f
¶
=  3 N
and F
y
y
= 
¶f
¶
=  4 N
PR
PQ
=
5
4
Þ PR PQ = ´
5
4
= 10 m
\ Work done by the conservative force on
the particle
= ´ F PR
net
= ´ 5 10 N m
= 50 Nm = 50 J
\ Correct option is (c).
14. Both at x x =
1
and x x =
2
the force acting
on the body is zero i.e., it is in equilibrium.
Now, if the body (when at x x =
1
) is moved
towards right (i.e., x x >
1
) the force acting
on it is + ive i.e., the body will not come
back and if the body (when at x x =
2
) is
moved toward rght (i.e., x x >
2
) the force
acting on it is  ive i.e., the body will
return back. Then,x x =
2
is the position of
stable equilibrium.
\ Correct option is (b).
130  Mechanics1
v
mg
u
mg
®
®
®
®
q
x
2x
mmg
m
v
T
®
®
P (6m, 8m)
4N
3N
F
net = 5N
x
R
y
6 m Q
Page 4
12. Decrease in KE = 
1
2
0
2
mv
=
1
2
2
mv
Change in mechanical energy
PE + KE =
1
2
2
mv
Decrease in ME is used up in doing work
against friction.
In this case the mechanical energy is
being used up in doing work against
friction and in increasing the PE of the
block.
\ Change ME = 
1
2
2
mv mgh
Thus, assertion is true.
As explained above the reason is false.
(m does not change with the increase in
angle of inclination)
\ Correct option is (c).
Objective Questions (Level 2)
Single Correct Option
1. Increase in KE of bead = Work done by
gravity + work done by force F
\
1
2
2
mv mgh FR = +
(Displacement of force F is R)
= ´ ´ + ´
1
2
10 5 5 5 = 50
\ u
m
=
100
= 200 =

14.14 ms
1
\ Correct option is (a).
2. P Fv =
= mav
=
æ
è
ç
ö
ø
÷
m v
dv
dx
v a
dv
dt
dv
ds
ds
dt
v
dv
ds
= = × =
æ
è
ç
ö
ø
÷
\ ds
m
p
v dv =
2
i.e., ds
m
p
v dv
v
v
ò ò
=
2
2
or s
m
p
v
v
v
=
é
ë
ê
ù
û
ú
3
2
3
= 
m
p
v v
3
2
3 3
[( ) ( ) ] =
7
3
3
mv
p
\ Correct option is (a).
3. Loss of PE of block = Gain in PE of spring
mg d k ( ) sin + ° = ´ 2 30
1
2
2
2
or 10 10 2
1
2
1
2
100 2
2
´ ´ + ´ = ´ ´ ( ) d
d = 2 m
Let, v
A
= velocity of block when it just
touches spring
\
1
2
30
2
mv mgd
A
= ° sin
v
A
2
10 2 2
1
2
= ´ ´ ´
v
A
=

20
1
ms
Correct option is (a).
128  Mechanics1
30°
f'
v'
Stops
PE = mgh
h
v
Slope
CaseI Rough
f
d
A
k
2m
A
30°
d
m
Work, Energy and Power  129
4. Mass per unit length of chain =
m
R p / 2
dm Rd
m
R
= q
p /2
y
ydm
dm
CM
=
ò
ò
=

ò
R
m
R
Rd
m
0
2
1
2
p
q
p
q
/
( cos )
/
= 
ò
2
1
0
2 R
d
p
q q
p
( cos )
/
= 
é
ë
ê
ù
û
ú
2
2
1
R
p
p
= 
é
ë
ê
ù
û
ú
R 1
2
p
Now,
1
2
2
mv mg y +
CM
or v g y
2
2 =
COM
v gR = 
æ
è
ç
ö
ø
÷ 2 1
2
p
\ Correct option is (c).
5. The moment string is cut
Net force on A mg = 2 (downward)
\ a g
1
2 =
Net force on B = 0
\ a
2
0 =
\ Correct option is (b).
6. Let x be the expansion in the spring.
Increase in PE of spring = Decrease in PE
of block C
1
2
2
1
kx M gx =
i.e., kx M g = 2
1
For block A to remain at rest
kx Mg = m
min
or 2
1
M g Mg = m
min
\ m
min
=
2
1
M
M
\ Correct option is (c).
7. T mg
i
=
KX
mg
mg
i
= =
2
2
When one spring is cut. It means KX
i
becomes zero. Downward acceleration,
a
KX
m
mg
m
g
i
= = =
2 2 2
Now drawing FBD of lower mass :
mg T m a
mg
f
 = = .
2
\ T
mg
f
=
2
or DT T T
mg
f i
=  =
2
8. a
mg mg
m
=
 sin cos q m q
=  g g sin cos q m q
For v to be maximum
dv
dx
= 0
or v
dv
dx
= 0
F (= mg)
mg
mg
A
B
F = mg
T = mg
T
F = kx = mg
mg (weight)
F
mg
Mg
1
A B
kx kx
T T
C
T
T
R (1 – cos q)
CM
y
( )
R 1 –
2
p
q
q
a
x
mmg cos q
mg sin q
or
dx
dt
dv
dx
× = 0
or
dv
dt
= 0
or a = 0
i.e., g g sin cos q m q =
or sin cos q m q =
or
3
5
3
10
4
5
= × x
Þ x = =
10
4
2.5 m
\ Correct option is (d).
9. (a) Between points E and F,
dU
dr
is  ive.
Now, F
dU
dr
=  , the force between E and F
will be + ive i.e., repulsive.
(b) At point C the potential energy is
minimum. Thus, C is point of stable
equilibrium.
\ Correct option is (c).
10. Power = ×
® ®
F v
= × +
® ® ®
m t g u g ( )
= ° + + ×
® ®
mg u m t cos ( ) 90 q g g
i.e., P mg u mg t =  + sin q
2
[   ,  u g
® ®
= = u g]
Therefore, the graph between P and t will
be as shown in option (c).
11.
PE of spring due to its compression by x
=  Work done by frictional force when
displaced by 2x
i.e.,
1
2
2
2
kx mg x = m
Þ x
mg
x
=
4 m
\ Correct option is (c).
12. Power delivered by man = ×
® ®
T v
= T vcos q
  T
®
= T
and   v
®
= T
13. f = + 3 4 x y
\ F
x
x
= 
¶f
¶
=  3 N
and F
y
y
= 
¶f
¶
=  4 N
PR
PQ
=
5
4
Þ PR PQ = ´
5
4
= 10 m
\ Work done by the conservative force on
the particle
= ´ F PR
net
= ´ 5 10 N m
= 50 Nm = 50 J
\ Correct option is (c).
14. Both at x x =
1
and x x =
2
the force acting
on the body is zero i.e., it is in equilibrium.
Now, if the body (when at x x =
1
) is moved
towards right (i.e., x x >
1
) the force acting
on it is + ive i.e., the body will not come
back and if the body (when at x x =
2
) is
moved toward rght (i.e., x x >
2
) the force
acting on it is  ive i.e., the body will
return back. Then,x x =
2
is the position of
stable equilibrium.
\ Correct option is (b).
130  Mechanics1
v
mg
u
mg
®
®
®
®
q
x
2x
mmg
m
v
T
®
®
P (6m, 8m)
4N
3N
F
net = 5N
x
R
y
6 m Q
15. The man will stop when
m q q mg mg cos sin =
or ( ) cos sin m q q
0
x mg mg =
or x =
tan q
m
0
16. Taking moment about A
AC F k x l × = ( )
2
\ AC
k x l
F
=
( )
2
=
+
( )
( )
k x l
k k x
2
1 2
=
+
k
k k
l
2
1 2
\ Correct option is (d).
17. mgh + work done by the force of friction
=
1
2
2
mv
\ Work done by the force of friction
= 
1
2
2
mv mgh
= ´ ´
æ
è
ç
ö
ø
÷  ´ ´
1
2
1 2 1 10 1
2
( )
=  8 J
\ Correct option is (c).
18. U
a
x
b
x
= 
12 6
U ax bx = 
  12 6
For stable equilibrium,
dU
dx
= 0
i.e., a x b x ( ) ( )    =
 
12 6 0
13 7
i.e., 6 12
7 12
b x a x
 
=
i.e., x
a
b
=
æ
è
ç
ö
ø
÷
2
1
6
\ Correct option is (a).
19. For the rotational equilibrium of the rod
m g
l
k x l × =
2
i.e., x
mg
k
=
2
\ PE stored in the spring =
1
2
2
kx
=
æ
è
ç
ö
ø
÷
1
2 2
2
k
mg
k
=
( ) mg
k
2
8
\ Correct option is (c).
20. a
m m
m m
g =

+
1 2
1 2
Speed (v) with which mass m
1
strikes the
floor
= + 0 2
2
gh =

+
æ
è
ç
ö
ø
÷ 2
1 2
1 2
m m
m m
gh
\ Correct option is (a).
21. F ax bx =  +
2
\  =  +
dU
dx
ax bx
2
or dU ax bx dx =  ( )
2
or U ax bx dx = 
ò
( )
2
or U
ax bx
c =  +
2 3
2 3
At x = 0, F = 0
\ U = 0
and so, c = 0
Thus, U
ax bx
= 
2 3
2 3
U = 0, when
ax bx
2 3
2 3
=
i.e., at x
a
b
=
3
2
Work, Energy and Power  131
C
B A
kx
1
kx
2
F = (k + k )x
12
q
x
mmg cos q
mg sin q
l kx
l/2
mg
Page 5
12. Decrease in KE = 
1
2
0
2
mv
=
1
2
2
mv
Change in mechanical energy
PE + KE =
1
2
2
mv
Decrease in ME is used up in doing work
against friction.
In this case the mechanical energy is
being used up in doing work against
friction and in increasing the PE of the
block.
\ Change ME = 
1
2
2
mv mgh
Thus, assertion is true.
As explained above the reason is false.
(m does not change with the increase in
angle of inclination)
\ Correct option is (c).
Objective Questions (Level 2)
Single Correct Option
1. Increase in KE of bead = Work done by
gravity + work done by force F
\
1
2
2
mv mgh FR = +
(Displacement of force F is R)
= ´ ´ + ´
1
2
10 5 5 5 = 50
\ u
m
=
100
= 200 =

14.14 ms
1
\ Correct option is (a).
2. P Fv =
= mav
=
æ
è
ç
ö
ø
÷
m v
dv
dx
v a
dv
dt
dv
ds
ds
dt
v
dv
ds
= = × =
æ
è
ç
ö
ø
÷
\ ds
m
p
v dv =
2
i.e., ds
m
p
v dv
v
v
ò ò
=
2
2
or s
m
p
v
v
v
=
é
ë
ê
ù
û
ú
3
2
3
= 
m
p
v v
3
2
3 3
[( ) ( ) ] =
7
3
3
mv
p
\ Correct option is (a).
3. Loss of PE of block = Gain in PE of spring
mg d k ( ) sin + ° = ´ 2 30
1
2
2
2
or 10 10 2
1
2
1
2
100 2
2
´ ´ + ´ = ´ ´ ( ) d
d = 2 m
Let, v
A
= velocity of block when it just
touches spring
\
1
2
30
2
mv mgd
A
= ° sin
v
A
2
10 2 2
1
2
= ´ ´ ´
v
A
=

20
1
ms
Correct option is (a).
128  Mechanics1
30°
f'
v'
Stops
PE = mgh
h
v
Slope
CaseI Rough
f
d
A
k
2m
A
30°
d
m
Work, Energy and Power  129
4. Mass per unit length of chain =
m
R p / 2
dm Rd
m
R
= q
p /2
y
ydm
dm
CM
=
ò
ò
=

ò
R
m
R
Rd
m
0
2
1
2
p
q
p
q
/
( cos )
/
= 
ò
2
1
0
2 R
d
p
q q
p
( cos )
/
= 
é
ë
ê
ù
û
ú
2
2
1
R
p
p
= 
é
ë
ê
ù
û
ú
R 1
2
p
Now,
1
2
2
mv mg y +
CM
or v g y
2
2 =
COM
v gR = 
æ
è
ç
ö
ø
÷ 2 1
2
p
\ Correct option is (c).
5. The moment string is cut
Net force on A mg = 2 (downward)
\ a g
1
2 =
Net force on B = 0
\ a
2
0 =
\ Correct option is (b).
6. Let x be the expansion in the spring.
Increase in PE of spring = Decrease in PE
of block C
1
2
2
1
kx M gx =
i.e., kx M g = 2
1
For block A to remain at rest
kx Mg = m
min
or 2
1
M g Mg = m
min
\ m
min
=
2
1
M
M
\ Correct option is (c).
7. T mg
i
=
KX
mg
mg
i
= =
2
2
When one spring is cut. It means KX
i
becomes zero. Downward acceleration,
a
KX
m
mg
m
g
i
= = =
2 2 2
Now drawing FBD of lower mass :
mg T m a
mg
f
 = = .
2
\ T
mg
f
=
2
or DT T T
mg
f i
=  =
2
8. a
mg mg
m
=
 sin cos q m q
=  g g sin cos q m q
For v to be maximum
dv
dx
= 0
or v
dv
dx
= 0
F (= mg)
mg
mg
A
B
F = mg
T = mg
T
F = kx = mg
mg (weight)
F
mg
Mg
1
A B
kx kx
T T
C
T
T
R (1 – cos q)
CM
y
( )
R 1 –
2
p
q
q
a
x
mmg cos q
mg sin q
or
dx
dt
dv
dx
× = 0
or
dv
dt
= 0
or a = 0
i.e., g g sin cos q m q =
or sin cos q m q =
or
3
5
3
10
4
5
= × x
Þ x = =
10
4
2.5 m
\ Correct option is (d).
9. (a) Between points E and F,
dU
dr
is  ive.
Now, F
dU
dr
=  , the force between E and F
will be + ive i.e., repulsive.
(b) At point C the potential energy is
minimum. Thus, C is point of stable
equilibrium.
\ Correct option is (c).
10. Power = ×
® ®
F v
= × +
® ® ®
m t g u g ( )
= ° + + ×
® ®
mg u m t cos ( ) 90 q g g
i.e., P mg u mg t =  + sin q
2
[   ,  u g
® ®
= = u g]
Therefore, the graph between P and t will
be as shown in option (c).
11.
PE of spring due to its compression by x
=  Work done by frictional force when
displaced by 2x
i.e.,
1
2
2
2
kx mg x = m
Þ x
mg
x
=
4 m
\ Correct option is (c).
12. Power delivered by man = ×
® ®
T v
= T vcos q
  T
®
= T
and   v
®
= T
13. f = + 3 4 x y
\ F
x
x
= 
¶f
¶
=  3 N
and F
y
y
= 
¶f
¶
=  4 N
PR
PQ
=
5
4
Þ PR PQ = ´
5
4
= 10 m
\ Work done by the conservative force on
the particle
= ´ F PR
net
= ´ 5 10 N m
= 50 Nm = 50 J
\ Correct option is (c).
14. Both at x x =
1
and x x =
2
the force acting
on the body is zero i.e., it is in equilibrium.
Now, if the body (when at x x =
1
) is moved
towards right (i.e., x x >
1
) the force acting
on it is + ive i.e., the body will not come
back and if the body (when at x x =
2
) is
moved toward rght (i.e., x x >
2
) the force
acting on it is  ive i.e., the body will
return back. Then,x x =
2
is the position of
stable equilibrium.
\ Correct option is (b).
130  Mechanics1
v
mg
u
mg
®
®
®
®
q
x
2x
mmg
m
v
T
®
®
P (6m, 8m)
4N
3N
F
net = 5N
x
R
y
6 m Q
15. The man will stop when
m q q mg mg cos sin =
or ( ) cos sin m q q
0
x mg mg =
or x =
tan q
m
0
16. Taking moment about A
AC F k x l × = ( )
2
\ AC
k x l
F
=
( )
2
=
+
( )
( )
k x l
k k x
2
1 2
=
+
k
k k
l
2
1 2
\ Correct option is (d).
17. mgh + work done by the force of friction
=
1
2
2
mv
\ Work done by the force of friction
= 
1
2
2
mv mgh
= ´ ´
æ
è
ç
ö
ø
÷  ´ ´
1
2
1 2 1 10 1
2
( )
=  8 J
\ Correct option is (c).
18. U
a
x
b
x
= 
12 6
U ax bx = 
  12 6
For stable equilibrium,
dU
dx
= 0
i.e., a x b x ( ) ( )    =
 
12 6 0
13 7
i.e., 6 12
7 12
b x a x
 
=
i.e., x
a
b
=
æ
è
ç
ö
ø
÷
2
1
6
\ Correct option is (a).
19. For the rotational equilibrium of the rod
m g
l
k x l × =
2
i.e., x
mg
k
=
2
\ PE stored in the spring =
1
2
2
kx
=
æ
è
ç
ö
ø
÷
1
2 2
2
k
mg
k
=
( ) mg
k
2
8
\ Correct option is (c).
20. a
m m
m m
g =

+
1 2
1 2
Speed (v) with which mass m
1
strikes the
floor
= + 0 2
2
gh =

+
æ
è
ç
ö
ø
÷ 2
1 2
1 2
m m
m m
gh
\ Correct option is (a).
21. F ax bx =  +
2
\  =  +
dU
dx
ax bx
2
or dU ax bx dx =  ( )
2
or U ax bx dx = 
ò
( )
2
or U
ax bx
c =  +
2 3
2 3
At x = 0, F = 0
\ U = 0
and so, c = 0
Thus, U
ax bx
= 
2 3
2 3
U = 0, when
ax bx
2 3
2 3
=
i.e., at x
a
b
=
3
2
Work, Energy and Power  131
C
B A
kx
1
kx
2
F = (k + k )x
12
q
x
mmg cos q
mg sin q
l kx
l/2
mg
dU
dx
= 0, when F = 0
i.e.,  + = ax bx
2
0
i.e., at x
a
b
=
Graph between U and x will be
\ Correct option is (c).
22.
W
W
Fs
Fs
A
B
= =
1
1
\ Correct option is (c).
W
W
mv
mv
A
B
A
B
= =
1
2
1
2
4
1
2
2
Þ
v
v
A
B
2
2
4
1
=
Þ
v
v
A
B
=
2
1
W
W
A
B
=
1
1
\
K
K
A
B
=
1
1
23. U
i
at (1, 1) = + = k k ( ) 1 1 2
U
f
at (2, 3) = + = k k ( ) 2 3 5
W U U
f i
= 
=  = 5 2 3 k k k
\ Correct option is (b).
24. Gain in PE of spring = Loss of PE of block
\
1
2
2
k x mg h x
max max
( ) = + …(i)
\ From above Eq. (i),
x
max
depends upon h and also x
max
depends upon k.
KE of the block will be maximum when it
is just at the point of touching the plank
and at this moment there would no
compression in the spring.
Maximum KE of block = mgh
\ Correct option is (c).
25. Gain in KE of chain = Decrease in PE of
chain
When the whole chain has justcome out of
the tube.
or
1
2
2
2
2
mv mg
r
= +
æ
è
ç
ö
ø
÷
p
p
p
\ v gr + +
æ
è
ç
ö
ø
÷ 2
2
2 p
p
\ Correct option is (b).
26. Acceleration of the block will decrease as
the block moves to the right and spring
expands the velocity (v) of block will be
maximum, when
1
2
1
2
2 2
mv kx =
132  Mechanics1
3a
2b
X a
b
O
x
max
h
PE = 0 level
for block
2r
p
CM (i)
CM (f)
pr
pr/2
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