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NEET Exam  >  NEET Notes  >  DC Pandey Solutions for NEET Physics  >  DC Pandey Solutions: Work, energy and Power- 1

DC Pandey Solutions: Work, energy and Power- 1 | DC Pandey Solutions for NEET Physics PDF Download

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 Page 1


Introductory Exercise 6.1
1. Work done by F
®
 = ×
® ®
F x
          | | ,| | F N
® ®
= = F N, | | W
®
= W
| | T
®
= T and | | x
®
= x
             =
® ®
| | | |cos F x p
             = - F x
Work done by | | N
®
 = ×
® ®
N x
             =
® ®
| | | |cos N x
p
2
             =0
Work done by W W x
® ® ®
= ×
     = ×
® ®
| || |cos W x
p
2
=0
Work done by T
®
 = ×
® ®
T x
              =
® ®
| | | |cos T x 0
              = Tx
2. Work done by F
®
 = ×
® ®
F l
              = ×
® ®
| || |cos F l p
              = - F l
Work done by N N l
® ® ®
= ×
       = × =
® ®
| || |cos N l
p
2
0
Work done by | | W W l
® ® ®
= ×
              = +
æ
è
ç
ö
ø
÷
® ®
| || |cos W l
p
a
2
              = -W l sina
Work done by | | T T l
® ® ®
= ×
        =
® ®
| || |cos T l b
     = T l cos b
6 Work, Energy and Power
T
N
F
W
– x
T
®
®
®
®
®
®
a
F
N
T
b
l
W
|F| = F, |N| = N
|W| = W, |T| = T
|l| = l
®
®
®
®
®
® ®
® ®
®
®
Page 2


Introductory Exercise 6.1
1. Work done by F
®
 = ×
® ®
F x
          | | ,| | F N
® ®
= = F N, | | W
®
= W
| | T
®
= T and | | x
®
= x
             =
® ®
| | | |cos F x p
             = - F x
Work done by | | N
®
 = ×
® ®
N x
             =
® ®
| | | |cos N x
p
2
             =0
Work done by W W x
® ® ®
= ×
     = ×
® ®
| || |cos W x
p
2
=0
Work done by T
®
 = ×
® ®
T x
              =
® ®
| | | |cos T x 0
              = Tx
2. Work done by F
®
 = ×
® ®
F l
              = ×
® ®
| || |cos F l p
              = - F l
Work done by N N l
® ® ®
= ×
       = × =
® ®
| || |cos N l
p
2
0
Work done by | | W W l
® ® ®
= ×
              = +
æ
è
ç
ö
ø
÷
® ®
| || |cos W l
p
a
2
              = -W l sina
Work done by | | T T l
® ® ®
= ×
        =
® ®
| || |cos T l b
     = T l cos b
6 Work, Energy and Power
T
N
F
W
– x
T
®
®
®
®
®
®
a
F
N
T
b
l
W
|F| = F, |N| = N
|W| = W, |T| = T
|l| = l
®
®
®
®
®
® ®
® ®
®
®
3. W T
g
® ®
®
- = m
4
| | g
®
= g and | | l
®
= l
\ | | T g
g
® ®
®
= - m
m
4
=
® 3
4
m g
Work done by string = ×
® ®
T l
              =
® ®
| || |cos T l p
              =
½
½
½ ½
½
½
® ®
3
4
mg l cosp
              = -
3
4
mgl
4.     m N F = ° cos45              …(i)
     N W = =
® ®
| |, | | N W
      s F = =
® ®
| |, | | s F , g =
®
| | g
    N F W + ° = sin45 …(ii)
Substituting value of N from Eq. (ii) in
Eq. (i).
m ( sin ) cos W F F - ° = ° 45 45
or    
1
4
1
2
1
2
W F F -
æ
è
ç
ö
ø
÷=
or          W
F F
- =
2
4
2
or          
5
2
F
W =
or F W mg = =
2
5
2
5
Work done by force F = ×
® ®
F s
           = °
® ®
| | | |cos F s 45 = F s
1
2
           =
æ
è
ç
ö
ø
÷
2
5
1
2
mg s =
mgs
5
           =
´ ´ 1.8 10 2
5
           =7.2 J
Work done by friction = ×
® ®
m N s
           =
® ®
m p | | | |cos N s
           = -m Ns
           = - ° F s cos45
           = -7.2 J
Work done by gravity = ×
® ®
g s
           = ×
® ®
| || |cos g s
p
2
           =0
5.          F mg = ° sin45
           = ´ ´ 1 10
1
2
           =5 2 N
Displacement of lift in 1s  = 2 m
Work, Energy and Power | 109
W
N
mN
s
F cos 45°
F F
F sin 45°
45°
45°
mg 
mg cos 45°
F
–1
2ms
sin 45°
W = mg
T
a =
g
4
l
®
®
®
Page 3


Introductory Exercise 6.1
1. Work done by F
®
 = ×
® ®
F x
          | | ,| | F N
® ®
= = F N, | | W
®
= W
| | T
®
= T and | | x
®
= x
             =
® ®
| | | |cos F x p
             = - F x
Work done by | | N
®
 = ×
® ®
N x
             =
® ®
| | | |cos N x
p
2
             =0
Work done by W W x
® ® ®
= ×
     = ×
® ®
| || |cos W x
p
2
=0
Work done by T
®
 = ×
® ®
T x
              =
® ®
| | | |cos T x 0
              = Tx
2. Work done by F
®
 = ×
® ®
F l
              = ×
® ®
| || |cos F l p
              = - F l
Work done by N N l
® ® ®
= ×
       = × =
® ®
| || |cos N l
p
2
0
Work done by | | W W l
® ® ®
= ×
              = +
æ
è
ç
ö
ø
÷
® ®
| || |cos W l
p
a
2
              = -W l sina
Work done by | | T T l
® ® ®
= ×
        =
® ®
| || |cos T l b
     = T l cos b
6 Work, Energy and Power
T
N
F
W
– x
T
®
®
®
®
®
®
a
F
N
T
b
l
W
|F| = F, |N| = N
|W| = W, |T| = T
|l| = l
®
®
®
®
®
® ®
® ®
®
®
3. W T
g
® ®
®
- = m
4
| | g
®
= g and | | l
®
= l
\ | | T g
g
® ®
®
= - m
m
4
=
® 3
4
m g
Work done by string = ×
® ®
T l
              =
® ®
| || |cos T l p
              =
½
½
½ ½
½
½
® ®
3
4
mg l cosp
              = -
3
4
mgl
4.     m N F = ° cos45              …(i)
     N W = =
® ®
| |, | | N W
      s F = =
® ®
| |, | | s F , g =
®
| | g
    N F W + ° = sin45 …(ii)
Substituting value of N from Eq. (ii) in
Eq. (i).
m ( sin ) cos W F F - ° = ° 45 45
or    
1
4
1
2
1
2
W F F -
æ
è
ç
ö
ø
÷=
or          W
F F
- =
2
4
2
or          
5
2
F
W =
or F W mg = =
2
5
2
5
Work done by force F = ×
® ®
F s
           = °
® ®
| | | |cos F s 45 = F s
1
2
           =
æ
è
ç
ö
ø
÷
2
5
1
2
mg s =
mgs
5
           =
´ ´ 1.8 10 2
5
           =7.2 J
Work done by friction = ×
® ®
m N s
           =
® ®
m p | | | |cos N s
           = -m Ns
           = - ° F s cos45
           = -7.2 J
Work done by gravity = ×
® ®
g s
           = ×
® ®
| || |cos g s
p
2
           =0
5.          F mg = ° sin45
           = ´ ´ 1 10
1
2
           =5 2 N
Displacement of lift in 1s  = 2 m
Work, Energy and Power | 109
W
N
mN
s
F cos 45°
F F
F sin 45°
45°
45°
mg 
mg cos 45°
F
–1
2ms
sin 45°
W = mg
T
a =
g
4
l
®
®
®
Work done  by force of friction ( ) F = ×
® ®
F s
      = °
® ®
| | | |cos F s 45
     = ° = × × = F scos 45 5 2 2
1
2
10 Nm
6. Total work-done by spring on both masses
= PE of the spring when stretched by 2
0
x
=
1
2
2
0
2
k x ( )
= 2
0
2
k x
\ Work done by spring on each mass 
=
2
2
0
2
kx
 =kx
0
2
 
7. Work done = Area under the curve
  = + + + A A A A
1 2 3 4
  =
- ´-
é
ë
ê
ù
û
ú
+
- ´
é
ë
ê
ù
û
ú
+ ´
2 10
2
2 10
2
10 2 [ ]
+
´
é
ë
ê
ù
û
ú
2 10
2
    =30 Nm
Introductory Exercise 6.2
1. a
s
=
-
= -
-
-
20
2
10
1
2
ms
ms
\ F ma = = ´ -
-
2 10
2
kg ms = - 20 N
s = Area under curve    
= ´ ´
-
1
2
25 20
1
ms   
=20 m
\ Work done = = - F s ( ) ( ) 20 20 N m
           = - 400 Nm
2. According to A (inertial frame)
Acceleration of P a =
\        Force on P ma = 
Work done over s displacement = mas
Now, v u as
2 2
2 = +
                =2as (Q u =0)
\ Gain in KE = = =
1
2
1
2
2
2
mv m as mas
\ DK W = (Work -Energy theorem)
According to B (non-inertial frame)
work done
  = Work done by F + Work done by f
p
 (pseudo force)
       = + - ( ) ( ) mas mas = 0
As P is at rest , DK = 0
\ DK W = (Work -Energy theorem)
Note In in er tial frames one has to also to con sider
work-done due to pseudo forces, while ap ply ing
Work-en ergy the o rem.
3.      v x =a
\     a
dv
dt x
dx
dt
= =a
1
2
       =a a
1
2 x
x =
a
2
2
     F ma m = =
a
2
2
\    W
m
b =
a
2
2
110 | Mechanics-1
m m
m m
x
0
2 4
– 4 – 2
A
1
A
2
A
3
A
4
10
– 10
x (m) 
F(N)
Q
P
a
A
m
Q
P F
a
B
f
P
Page 4


Introductory Exercise 6.1
1. Work done by F
®
 = ×
® ®
F x
          | | ,| | F N
® ®
= = F N, | | W
®
= W
| | T
®
= T and | | x
®
= x
             =
® ®
| | | |cos F x p
             = - F x
Work done by | | N
®
 = ×
® ®
N x
             =
® ®
| | | |cos N x
p
2
             =0
Work done by W W x
® ® ®
= ×
     = ×
® ®
| || |cos W x
p
2
=0
Work done by T
®
 = ×
® ®
T x
              =
® ®
| | | |cos T x 0
              = Tx
2. Work done by F
®
 = ×
® ®
F l
              = ×
® ®
| || |cos F l p
              = - F l
Work done by N N l
® ® ®
= ×
       = × =
® ®
| || |cos N l
p
2
0
Work done by | | W W l
® ® ®
= ×
              = +
æ
è
ç
ö
ø
÷
® ®
| || |cos W l
p
a
2
              = -W l sina
Work done by | | T T l
® ® ®
= ×
        =
® ®
| || |cos T l b
     = T l cos b
6 Work, Energy and Power
T
N
F
W
– x
T
®
®
®
®
®
®
a
F
N
T
b
l
W
|F| = F, |N| = N
|W| = W, |T| = T
|l| = l
®
®
®
®
®
® ®
® ®
®
®
3. W T
g
® ®
®
- = m
4
| | g
®
= g and | | l
®
= l
\ | | T g
g
® ®
®
= - m
m
4
=
® 3
4
m g
Work done by string = ×
® ®
T l
              =
® ®
| || |cos T l p
              =
½
½
½ ½
½
½
® ®
3
4
mg l cosp
              = -
3
4
mgl
4.     m N F = ° cos45              …(i)
     N W = =
® ®
| |, | | N W
      s F = =
® ®
| |, | | s F , g =
®
| | g
    N F W + ° = sin45 …(ii)
Substituting value of N from Eq. (ii) in
Eq. (i).
m ( sin ) cos W F F - ° = ° 45 45
or    
1
4
1
2
1
2
W F F -
æ
è
ç
ö
ø
÷=
or          W
F F
- =
2
4
2
or          
5
2
F
W =
or F W mg = =
2
5
2
5
Work done by force F = ×
® ®
F s
           = °
® ®
| | | |cos F s 45 = F s
1
2
           =
æ
è
ç
ö
ø
÷
2
5
1
2
mg s =
mgs
5
           =
´ ´ 1.8 10 2
5
           =7.2 J
Work done by friction = ×
® ®
m N s
           =
® ®
m p | | | |cos N s
           = -m Ns
           = - ° F s cos45
           = -7.2 J
Work done by gravity = ×
® ®
g s
           = ×
® ®
| || |cos g s
p
2
           =0
5.          F mg = ° sin45
           = ´ ´ 1 10
1
2
           =5 2 N
Displacement of lift in 1s  = 2 m
Work, Energy and Power | 109
W
N
mN
s
F cos 45°
F F
F sin 45°
45°
45°
mg 
mg cos 45°
F
–1
2ms
sin 45°
W = mg
T
a =
g
4
l
®
®
®
Work done  by force of friction ( ) F = ×
® ®
F s
      = °
® ®
| | | |cos F s 45
     = ° = × × = F scos 45 5 2 2
1
2
10 Nm
6. Total work-done by spring on both masses
= PE of the spring when stretched by 2
0
x
=
1
2
2
0
2
k x ( )
= 2
0
2
k x
\ Work done by spring on each mass 
=
2
2
0
2
kx
 =kx
0
2
 
7. Work done = Area under the curve
  = + + + A A A A
1 2 3 4
  =
- ´-
é
ë
ê
ù
û
ú
+
- ´
é
ë
ê
ù
û
ú
+ ´
2 10
2
2 10
2
10 2 [ ]
+
´
é
ë
ê
ù
û
ú
2 10
2
    =30 Nm
Introductory Exercise 6.2
1. a
s
=
-
= -
-
-
20
2
10
1
2
ms
ms
\ F ma = = ´ -
-
2 10
2
kg ms = - 20 N
s = Area under curve    
= ´ ´
-
1
2
25 20
1
ms   
=20 m
\ Work done = = - F s ( ) ( ) 20 20 N m
           = - 400 Nm
2. According to A (inertial frame)
Acceleration of P a =
\        Force on P ma = 
Work done over s displacement = mas
Now, v u as
2 2
2 = +
                =2as (Q u =0)
\ Gain in KE = = =
1
2
1
2
2
2
mv m as mas
\ DK W = (Work -Energy theorem)
According to B (non-inertial frame)
work done
  = Work done by F + Work done by f
p
 (pseudo force)
       = + - ( ) ( ) mas mas = 0
As P is at rest , DK = 0
\ DK W = (Work -Energy theorem)
Note In in er tial frames one has to also to con sider
work-done due to pseudo forces, while ap ply ing
Work-en ergy the o rem.
3.      v x =a
\     a
dv
dt x
dx
dt
= =a
1
2
       =a a
1
2 x
x =
a
2
2
     F ma m = =
a
2
2
\    W
m
b =
a
2
2
110 | Mechanics-1
m m
m m
x
0
2 4
– 4 – 2
A
1
A
2
A
3
A
4
10
– 10
x (m) 
F(N)
Q
P
a
A
m
Q
P F
a
B
f
P
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
 
Work, Energy and Power | 111
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


Introductory Exercise 6.1
1. Work done by F
®
 = ×
® ®
F x
          | | ,| | F N
® ®
= = F N, | | W
®
= W
| | T
®
= T and | | x
®
= x
             =
® ®
| | | |cos F x p
             = - F x
Work done by | | N
®
 = ×
® ®
N x
             =
® ®
| | | |cos N x
p
2
             =0
Work done by W W x
® ® ®
= ×
     = ×
® ®
| || |cos W x
p
2
=0
Work done by T
®
 = ×
® ®
T x
              =
® ®
| | | |cos T x 0
              = Tx
2. Work done by F
®
 = ×
® ®
F l
              = ×
® ®
| || |cos F l p
              = - F l
Work done by N N l
® ® ®
= ×
       = × =
® ®
| || |cos N l
p
2
0
Work done by | | W W l
® ® ®
= ×
              = +
æ
è
ç
ö
ø
÷
® ®
| || |cos W l
p
a
2
              = -W l sina
Work done by | | T T l
® ® ®
= ×
        =
® ®
| || |cos T l b
     = T l cos b
6 Work, Energy and Power
T
N
F
W
– x
T
®
®
®
®
®
®
a
F
N
T
b
l
W
|F| = F, |N| = N
|W| = W, |T| = T
|l| = l
®
®
®
®
®
® ®
® ®
®
®
3. W T
g
® ®
®
- = m
4
| | g
®
= g and | | l
®
= l
\ | | T g
g
® ®
®
= - m
m
4
=
® 3
4
m g
Work done by string = ×
® ®
T l
              =
® ®
| || |cos T l p
              =
½
½
½ ½
½
½
® ®
3
4
mg l cosp
              = -
3
4
mgl
4.     m N F = ° cos45              …(i)
     N W = =
® ®
| |, | | N W
      s F = =
® ®
| |, | | s F , g =
®
| | g
    N F W + ° = sin45 …(ii)
Substituting value of N from Eq. (ii) in
Eq. (i).
m ( sin ) cos W F F - ° = ° 45 45
or    
1
4
1
2
1
2
W F F -
æ
è
ç
ö
ø
÷=
or          W
F F
- =
2
4
2
or          
5
2
F
W =
or F W mg = =
2
5
2
5
Work done by force F = ×
® ®
F s
           = °
® ®
| | | |cos F s 45 = F s
1
2
           =
æ
è
ç
ö
ø
÷
2
5
1
2
mg s =
mgs
5
           =
´ ´ 1.8 10 2
5
           =7.2 J
Work done by friction = ×
® ®
m N s
           =
® ®
m p | | | |cos N s
           = -m Ns
           = - ° F s cos45
           = -7.2 J
Work done by gravity = ×
® ®
g s
           = ×
® ®
| || |cos g s
p
2
           =0
5.          F mg = ° sin45
           = ´ ´ 1 10
1
2
           =5 2 N
Displacement of lift in 1s  = 2 m
Work, Energy and Power | 109
W
N
mN
s
F cos 45°
F F
F sin 45°
45°
45°
mg 
mg cos 45°
F
–1
2ms
sin 45°
W = mg
T
a =
g
4
l
®
®
®
Work done  by force of friction ( ) F = ×
® ®
F s
      = °
® ®
| | | |cos F s 45
     = ° = × × = F scos 45 5 2 2
1
2
10 Nm
6. Total work-done by spring on both masses
= PE of the spring when stretched by 2
0
x
=
1
2
2
0
2
k x ( )
= 2
0
2
k x
\ Work done by spring on each mass 
=
2
2
0
2
kx
 =kx
0
2
 
7. Work done = Area under the curve
  = + + + A A A A
1 2 3 4
  =
- ´-
é
ë
ê
ù
û
ú
+
- ´
é
ë
ê
ù
û
ú
+ ´
2 10
2
2 10
2
10 2 [ ]
+
´
é
ë
ê
ù
û
ú
2 10
2
    =30 Nm
Introductory Exercise 6.2
1. a
s
=
-
= -
-
-
20
2
10
1
2
ms
ms
\ F ma = = ´ -
-
2 10
2
kg ms = - 20 N
s = Area under curve    
= ´ ´
-
1
2
25 20
1
ms   
=20 m
\ Work done = = - F s ( ) ( ) 20 20 N m
           = - 400 Nm
2. According to A (inertial frame)
Acceleration of P a =
\        Force on P ma = 
Work done over s displacement = mas
Now, v u as
2 2
2 = +
                =2as (Q u =0)
\ Gain in KE = = =
1
2
1
2
2
2
mv m as mas
\ DK W = (Work -Energy theorem)
According to B (non-inertial frame)
work done
  = Work done by F + Work done by f
p
 (pseudo force)
       = + - ( ) ( ) mas mas = 0
As P is at rest , DK = 0
\ DK W = (Work -Energy theorem)
Note In in er tial frames one has to also to con sider
work-done due to pseudo forces, while ap ply ing
Work-en ergy the o rem.
3.      v x =a
\     a
dv
dt x
dx
dt
= =a
1
2
       =a a
1
2 x
x =
a
2
2
     F ma m = =
a
2
2
\    W
m
b =
a
2
2
110 | Mechanics-1
m m
m m
x
0
2 4
– 4 – 2
A
1
A
2
A
3
A
4
10
– 10
x (m) 
F(N)
Q
P
a
A
m
Q
P F
a
B
f
P
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
 
Work, Energy and Power | 111
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
112 | Mechanics-1
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 and Power- 1 - DC Pandey Solutions for NEET Physics

1. What is work and how is it related to energy and power?
Ans. Work is defined as the transfer of energy that occurs when a force is applied to an object and it causes the object to move in the direction of the force. It is directly related to energy and power. The work done on an object is equal to the change in its energy. Power, on the other hand, is the rate at which work is done or energy is transferred. It is calculated by dividing the work done by the time taken.
2. How is work calculated and what are its units?
Ans. Work is calculated by multiplying the force applied to an object by the distance over which the force is applied. Mathematically, work (W) = force (F) × distance (d) × cosine(theta), where theta is the angle between the direction of the force and the direction of displacement. The SI unit of work is the joule (J), which is equal to one newton-meter (Nm).
3. What is the principle of conservation of energy and how does it relate to work?
Ans. The principle of conservation of energy states that energy cannot be created or destroyed, but it can only be transferred or transformed from one form to another. This principle is closely related to work because the work done on an object is equal to the change in its energy. In other words, the total energy of a system remains constant if no external forces are acting on it. Therefore, the work done on an object can either increase or decrease its energy, but the total energy of the system remains conserved.
4. How can power be calculated and what are its units?
Ans. Power is calculated by dividing the work done by the time taken. Mathematically, power (P) = work (W) / time (t). The SI unit of power is the watt (W), which is equal to one joule per second (J/s). In practical terms, power is a measure of how quickly work is done or energy is transferred.
5. What are some examples of work, energy, and power in everyday life?
Ans. Some examples of work, energy, and power in everyday life include: - Lifting a book from the ground to a shelf: This requires the application of a force (work) to move the book against gravity, which increases its potential energy. - Using a hairdryer: The electrical energy is converted into heat and kinetic energy, which requires power. - Riding a bicycle: The pedaling motion requires the application of a force (work) to move the bicycle, converting the rider's energy into kinetic energy. - Turning on a light bulb: The electrical energy is converted into light and heat energy, which requires power. - Using a computer: The electrical energy is converted into various forms of energy, such as light, heat, and sound, which requires power.
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Document Description: DC Pandey Solutions: Work, energy and Power- 1 for NEET 2024 is part of DC Pandey Solutions for NEET Physics preparation. The notes and questions for DC Pandey Solutions: Work, energy and Power- 1 have been prepared according to the NEET exam syllabus. Information about DC Pandey Solutions: Work, energy and Power- 1 covers topics like and DC Pandey Solutions: Work, energy and Power- 1 Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for DC Pandey Solutions: Work, energy and Power- 1.
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