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# Chapter 6 - Work, energy and Power (Part - 1) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

## DC Pandey (Questions & Solutions) of Physics: NEET

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## NEET : Chapter 6 - Work, energy and Power (Part - 1) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

``` 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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