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4. DK = Work done by F 
+ Work done by gravity
       = × × + × × 80 4 0 5 4 cos cos g p
       = + - 320 200 ( )
or       K K
f i
- =120 J
or           K
f
=120 J (as K
i
= 0)
5. Change in KE = Work done
1
2
1
2
mv mgR mgR = - + ( cos ) sin q q
Þ   v gR = - + 2 1 ( cos sin ) q q
6. DK W =
or   0
1
2
0
2
0
- = -
ò
mv Ax dx
x
or    - = -
1
2 2
0
2
2
mv A
x
Þ          x v
m
A
=
0
7. (a) If T mg = , the block will not get
ac cel er ated to gain KE. The value of T
must be greater that Mg.
\  Ans. False
(b) As some negative work  will be done by 
Mg, the work done by T will be more
that 40 J.
\  Ans. False
(c) Pulling force F will always be equal to 
T, as T is there only because of pulling.
\  Ans. True
(d) Work done by gravity will be negative
Ans. False
Introductory Exercise 6.3
1. In Fig. 1 Spring is having its natural length.
In Fig. 2 A is released. A goes down byx .
Spring get extended by x. Decrease in PE
of A is stored in spring as its PE.
\ mAg x k x =
1
2
2
Now, for the block B to just leave contact
with ground
kx mg =
i.e., 2m g mg
A
=   
Þ m
m
A
=
2
 
  
F = 80 N
5 g
4m
g
F
mg
ma = mg
R(1 – cosq)
O
R sin q
R
mg
mg
q
T T
T F
Hand
Mg
Fig. 1                         Fig. 2
B
A
m
Ground
x
B
A
m
Ground
mg
A
T
T
T
T
T
T
mg
Page 5


4. DK = Work done by F 
+ Work done by gravity
       = × × + × × 80 4 0 5 4 cos cos g p
       = + - 320 200 ( )
or       K K
f i
- =120 J
or           K
f
=120 J (as K
i
= 0)
5. Change in KE = Work done
1
2
1
2
mv mgR mgR = - + ( cos ) sin q q
Þ   v gR = - + 2 1 ( cos sin ) q q
6. DK W =
or   0
1
2
0
2
0
- = -
ò
mv Ax dx
x
or    - = -
1
2 2
0
2
2
mv A
x
Þ          x v
m
A
=
0
7. (a) If T mg = , the block will not get
ac cel er ated to gain KE. The value of T
must be greater that Mg.
\  Ans. False
(b) As some negative work  will be done by 
Mg, the work done by T will be more
that 40 J.
\  Ans. False
(c) Pulling force F will always be equal to 
T, as T is there only because of pulling.
\  Ans. True
(d) Work done by gravity will be negative
Ans. False
Introductory Exercise 6.3
1. In Fig. 1 Spring is having its natural length.
In Fig. 2 A is released. A goes down byx .
Spring get extended by x. Decrease in PE
of A is stored in spring as its PE.
\ mAg x k x =
1
2
2
Now, for the block B to just leave contact
with ground
kx mg =
i.e., 2m g mg
A
=   
Þ m
m
A
=
2
 
  
F = 80 N
5 g
4m
g
F
mg
ma = mg
R(1 – cosq)
O
R sin q
R
mg
mg
q
T T
T F
Hand
Mg
Fig. 1                         Fig. 2
B
A
m
Ground
x
B
A
m
Ground
mg
A
T
T
T
T
T
T
mg
2. Decrease in PE = mg
l
2
\         
1
2 2
2
mv mg
l
=  
i.e.,            v g l =
3. OA = 50 cm
\  Extension in spring (when collar is at A)
                  = 50 cm - 10 cm = 0.4 m         
Extension in spring (collar is at B)
= 30 cm - 10 cm
= 20 cm = 0.2 m 
KE of collar at B 
    = - PEofspring PEofspring
(collarat ) (collarat ) A B
             = ´ ´ -
1
2
2 2
K [( ) ( ) ] 0.4 0.2
or  
1 1
2
500 1
2
m
mv
B
= ´ ´ 2
or    v
B
=
´ 500 012
10
.
 =245 . s
-1
Extension in spring (collar arrives at  C)
          = + - [ ( ) ( ) ] 30 20 10
2 2
 cm = 0.26 m
KE of collar at C
= PE of spring - PE of spring 
(Collar at A)            (Collar at C)
                 = ´ ´ -
1
2
500 0 4 026
2 2
[( . ) ( . ) ]
or  
1
2
1
2
500 00924
2
mv
C
= ´ ´ .           
or      v
c
= ´
500
100
0.0924 =
-
2.15ms
1
4. Work done by man = +
m
gh Mgh
2
          = +
æ
è
ç
ö
ø
÷
m
M gh
2
5. When block of man M goes down by x, the
spring gets extended by x. Decrease in PE
of man M is stored in spring as its PE.
\ Mgx k x =
1
2
2
or   kx Mg = 2
For the block of man m to just slide
     kx mg mg = ° + ° sin cos 37 37 m
or     2
3
5
3
4
4
5
Mg mg mg = +
or      M m =
3
5
Introductory Exercise 6.4
1. Velocity at time t = 2 s
     v gt = = ´ =
-
10 2 20
1
ms
       Power = Force ´ velocity
            = mgv = ´ ´ 1 10 20 =200 W
2. Velocity at time = a t = ×
F
m
t
\ v
Ft
m
av
=
2
 (acceleration being constant)
  P F v
av av
= ´ =
F t
m
2
2
  
A
C 20 cm B 40 cm
30 cm
O
m
m
37°
37°
mg
T
T
T
T
T
T
x
mg sin 37°
mmg cos 37°
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FAQs on DC Pandey Solutions: Work, Energy & Power - 2 - Physics Class 11 - NEET

1. What is the definition of work in the context of physics?
Ans. In physics, work is defined as the product of the force applied to an object and the distance over which the force is applied. It is measured in joules (J) and can be positive or negative depending on the direction of the force and the displacement of the object.
2. How is energy related to work?
Ans. Energy is the capacity to do work. When work is done on an object, energy is transferred to that object, causing it to gain or lose energy. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.
3. What is the difference between work and power?
Ans. Work and power are related concepts but are not the same. Work is the transfer of energy that occurs when a force is applied to an object and the object is displaced. Power, on the other hand, is the rate at which work is done or the amount of work done per unit time. It is measured in watts (W).
4. How can we calculate the work done by a force?
Ans. The work done by a force can be calculated using the formula: work = force × displacement × cos(theta), where force is the magnitude of the force applied, displacement is the distance over which the force is applied, and theta is the angle between the force and the displacement vectors.
5. What factors affect the amount of work done?
Ans. The amount of work done depends on the magnitude of the force applied, the displacement of the object, and the angle between the force and the displacement vectors. If the force and displacement are in the same direction, more work is done, whereas if they are in opposite directions, less work is done. Additionally, the work done is zero if the force is perpendicular to the displacement.
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