Work, Energy and Power (Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Work, Energy and Power (Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

 Page 4


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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