DC Pandey Solutions: Work, energy and Power- 3

# DC Pandey Solutions: Work, energy and Power- 3 | DC Pandey Solutions for NEET Physics PDF Download

``` 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 | Mechanics-1
30°
f'
v'
Stops
PE = mgh
h
v
Slope
Case-I 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 | Mechanics-1
30°
f'
v'
Stops
PE = mgh
h
v
Slope
Case-I 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 | Mechanics-1
30°
f'
v'
Stops
PE = mgh
h
v
Slope
Case-I 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 | Mechanics-1
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 | Mechanics-1
30°
f'
v'
Stops
PE = mgh
h
v
Slope
Case-I 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 | Mechanics-1
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
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 | Mechanics-1
30°
f'
v'
Stops
PE = mgh
h
v
Slope
Case-I 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 | Mechanics-1
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
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 | Mechanics-1
3a
2b
X a
b
O
x
max
h
PE = 0 level
for block
2r
p
CM (i)
CM (f)
pr
pr/2
```

122 docs

## FAQs on DC Pandey Solutions: Work, energy and Power- 3 - DC Pandey Solutions for NEET Physics

 1. What is work in the context of physics?
Ans. Work in physics refers to the transfer of energy that occurs when an object is moved against an external force. It is calculated as the product of the force applied on the object and the displacement of the object in the direction of the force.
 2. How is power defined in physics?
Ans. Power in physics is defined as the rate at which work is done or energy is transferred. It is calculated as the work done divided by the time taken to do the work. The unit of power is the watt (W).
 3. What is the relation between work and energy?
Ans. Work and energy are closely related in physics. Work is the transfer of energy, and energy is the ability to do work. When work is done on an object, its energy changes. Similarly, when work is done by an object, it loses energy.
 4. How do you calculate the amount of work done?
Ans. The amount of work done is calculated by multiplying the force applied on an object by the displacement of the object in the direction of the force. Mathematically, work (W) = force (F) × displacement (d) × cos(θ), where θ is the angle between the force and displacement vectors.
 5. What is the work-energy theorem?
Ans. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. This theorem helps in relating the work done on an object to its resulting change in speed or motion. It is expressed as W = ΔKE, where W is the work done and ΔKE is the change in kinetic energy.

## DC Pandey Solutions for NEET Physics

122 docs

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