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# HC Verma Solution - Work & Energy Class 11 Notes | EduRev

## Class 11 : HC Verma Solution - Work & Energy Class 11 Notes | EduRev

``` Page 1

8.1
SOLUTIONS TO CONCEPTS
CHAPTER – 8
1. M = m
c
+ m
b
= 90kg
u =  6 km/h = 1.666 m/sec
v = 12 km/h = 3.333 m/sec
Increase in K.E. = ½ Mv
2
– ½ Mu
2
= ½ 90 × (3.333)
2
– ½ × 90 × (1.66)
2
= 494.5 – 124.6 = 374.8 ? 375 J
2. m
b
= 2 kg.
u = 10 m/sec
a = 3 m/aec
2
t = 5 sec
v = u + at = 10 + 3 I 5 = 25 m/sec.
?F.K.E = ½ mv
2
= ½ × 2 × 625 = 625 J.
3. F = 100 N
S = 4m, ? = 0°
? = S . F
? ?
???100 × 4 = 400 J ?
4. m = 5 kg
? = 30°
S = 10 m
F = mg
So, work done by the force of gravity
? = mgh = 5 × 9.8 × 5 = 245 J ?
5. F= 2.50N, S = 2.5m,  m =15g = 0.015kg.
So, w = F × S ? a =
m
F
=
015 . 0
5 . 2
=
3
500
m/s
2
=F × S cos 0° (acting along the same line)
= 2.5 × 2.5 = 6.25J
Let the velocity of the body at b = U. Applying work-energy principle ½ mv
2
– 0 = 6.25
? V =
015 . 0
2 25 . 6 ?
= 28.86 m/sec.
So, time taken to travel from A to B.
? t =
a
u v ?
=
500
3 86 . 28 ?
? Average power =
t
W
=
3 ) 86 . 28 (
500 25 . 6
?
?
= 36.1
6. Given
j
ˆ
3 i
ˆ
2 r
1
? ?
?
j
ˆ
2 i
ˆ
3 r
2
? ?
So, displacement vector is given by,
2 1
r r r
? ? ?
? ? ? j
ˆ
i
ˆ
) j
ˆ
3 i
ˆ
2 ( ) j
ˆ
2 i
ˆ
3 ( r ? ? ? ? ? ?
?
u=1.66 m/s ?
90kg ? 90kg ?
v=3.33 m/s ?
u=10 m/s ?
2 kg ? a
?
= 3m/s
2
F ?
4m ?
R ?
100 N ?
mg ?
30°
?
5 ?
mg ?
F ?
5 log ?
10m ?
30°
?
v ? ?
A ? B ?
Page 2

8.1
SOLUTIONS TO CONCEPTS
CHAPTER – 8
1. M = m
c
+ m
b
= 90kg
u =  6 km/h = 1.666 m/sec
v = 12 km/h = 3.333 m/sec
Increase in K.E. = ½ Mv
2
– ½ Mu
2
= ½ 90 × (3.333)
2
– ½ × 90 × (1.66)
2
= 494.5 – 124.6 = 374.8 ? 375 J
2. m
b
= 2 kg.
u = 10 m/sec
a = 3 m/aec
2
t = 5 sec
v = u + at = 10 + 3 I 5 = 25 m/sec.
?F.K.E = ½ mv
2
= ½ × 2 × 625 = 625 J.
3. F = 100 N
S = 4m, ? = 0°
? = S . F
? ?
???100 × 4 = 400 J ?
4. m = 5 kg
? = 30°
S = 10 m
F = mg
So, work done by the force of gravity
? = mgh = 5 × 9.8 × 5 = 245 J ?
5. F= 2.50N, S = 2.5m,  m =15g = 0.015kg.
So, w = F × S ? a =
m
F
=
015 . 0
5 . 2
=
3
500
m/s
2
=F × S cos 0° (acting along the same line)
= 2.5 × 2.5 = 6.25J
Let the velocity of the body at b = U. Applying work-energy principle ½ mv
2
– 0 = 6.25
? V =
015 . 0
2 25 . 6 ?
= 28.86 m/sec.
So, time taken to travel from A to B.
? t =
a
u v ?
=
500
3 86 . 28 ?
? Average power =
t
W
=
3 ) 86 . 28 (
500 25 . 6
?
?
= 36.1
6. Given
j
ˆ
3 i
ˆ
2 r
1
? ?
?
j
ˆ
2 i
ˆ
3 r
2
? ?
So, displacement vector is given by,
2 1
r r r
? ? ?
? ? ? j
ˆ
i
ˆ
) j
ˆ
3 i
ˆ
2 ( ) j
ˆ
2 i
ˆ
3 ( r ? ? ? ? ? ?
?
u=1.66 m/s ?
90kg ? 90kg ?
v=3.33 m/s ?
u=10 m/s ?
2 kg ? a
?
= 3m/s
2
F ?
4m ?
R ?
100 N ?
mg ?
30°
?
5 ?
mg ?
F ?
5 log ?
10m ?
30°
?
v ? ?
A ? B ?
Chapter 8
8.2
So, work done = s F
?
?
? = 5 × 1 + 5(-1) = 0
7. m
b
= 2kg, s = 40m, a = 0.5m/sec
2
So, force applied by the man on the box
F = m
b
a = 2 × (0.5) = 1 N
? = FS = 1 × 40 = 40 J ?
8. Given that F= a + bx
Where a and b are constants.
So, work done by this force during this force during the displacement x = 0 and  x = d is given
by
W =
? ?
? ?
d
0
d
0
dx ) bx a ( dx F = ax + (bx
2
/2) = [a + ½ bd] d
9. m
b
= 250g = .250 kg
? = 37°, S = 1m.
Frictional force f = ?R
mg sin ? = ? R ..(1)
mg cos ?? ..(2)
so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J
10. a =
) m M ( 2
F
?
(given)
a) from fig (1)
ma = ?
k
R
1
and R
1
= mg
? ? =
1
R
ma
=
g ) m M ( 2
F
?
b) Frictional force acting on the smaller block f = ?R =
) m M ( 2
F m
mg
g ) m M ( 2
F
?
?
? ?
?
c) Work done w = fs s = d
w = d
) m M ( 2
mF
?
?
=
) m M ( 2
mFd
?
?
11. Weight = 2000 N, S = 20m, ? = 0.2
a) R + Psin ? - 2000 = 0 ..(1)
P cos ? - 0.2 R =0 ..(2)
From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0
P =
? ? ? sin 2 . 0 cos
400
..(3)
So, work done by the person, W = PS cos ? =
? ? ?
?
sin 2 . 0 cos
cos 8000
=
? ? sin 2 . 0 1
8000
=
? ? tan 5
40000
b) For minimum magnitude of force from equn(1)
d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2
putting the value in equn (3)
W =
? ? tan 5
40000
=
) 2 . 5 (
40000
= 7690 J
12. w = 100 N,  ? = 37°,  s = 2m
R ?
m b g ?
m b a ? F ?
R ?
?R ?
1 m ?
mg ?
37° ?
M
?
F ?
m
?
R 1 ?
? k R 1 ?
ma ?
mg ?
?R 1 ?
R 2 ?
f
?
ma ?
mg ?
?R 2 ?
? ?
R
?
P
?
0.2R ?
2000 N ?
Page 3

8.1
SOLUTIONS TO CONCEPTS
CHAPTER – 8
1. M = m
c
+ m
b
= 90kg
u =  6 km/h = 1.666 m/sec
v = 12 km/h = 3.333 m/sec
Increase in K.E. = ½ Mv
2
– ½ Mu
2
= ½ 90 × (3.333)
2
– ½ × 90 × (1.66)
2
= 494.5 – 124.6 = 374.8 ? 375 J
2. m
b
= 2 kg.
u = 10 m/sec
a = 3 m/aec
2
t = 5 sec
v = u + at = 10 + 3 I 5 = 25 m/sec.
?F.K.E = ½ mv
2
= ½ × 2 × 625 = 625 J.
3. F = 100 N
S = 4m, ? = 0°
? = S . F
? ?
???100 × 4 = 400 J ?
4. m = 5 kg
? = 30°
S = 10 m
F = mg
So, work done by the force of gravity
? = mgh = 5 × 9.8 × 5 = 245 J ?
5. F= 2.50N, S = 2.5m,  m =15g = 0.015kg.
So, w = F × S ? a =
m
F
=
015 . 0
5 . 2
=
3
500
m/s
2
=F × S cos 0° (acting along the same line)
= 2.5 × 2.5 = 6.25J
Let the velocity of the body at b = U. Applying work-energy principle ½ mv
2
– 0 = 6.25
? V =
015 . 0
2 25 . 6 ?
= 28.86 m/sec.
So, time taken to travel from A to B.
? t =
a
u v ?
=
500
3 86 . 28 ?
? Average power =
t
W
=
3 ) 86 . 28 (
500 25 . 6
?
?
= 36.1
6. Given
j
ˆ
3 i
ˆ
2 r
1
? ?
?
j
ˆ
2 i
ˆ
3 r
2
? ?
So, displacement vector is given by,
2 1
r r r
? ? ?
? ? ? j
ˆ
i
ˆ
) j
ˆ
3 i
ˆ
2 ( ) j
ˆ
2 i
ˆ
3 ( r ? ? ? ? ? ?
?
u=1.66 m/s ?
90kg ? 90kg ?
v=3.33 m/s ?
u=10 m/s ?
2 kg ? a
?
= 3m/s
2
F ?
4m ?
R ?
100 N ?
mg ?
30°
?
5 ?
mg ?
F ?
5 log ?
10m ?
30°
?
v ? ?
A ? B ?
Chapter 8
8.2
So, work done = s F
?
?
? = 5 × 1 + 5(-1) = 0
7. m
b
= 2kg, s = 40m, a = 0.5m/sec
2
So, force applied by the man on the box
F = m
b
a = 2 × (0.5) = 1 N
? = FS = 1 × 40 = 40 J ?
8. Given that F= a + bx
Where a and b are constants.
So, work done by this force during this force during the displacement x = 0 and  x = d is given
by
W =
? ?
? ?
d
0
d
0
dx ) bx a ( dx F = ax + (bx
2
/2) = [a + ½ bd] d
9. m
b
= 250g = .250 kg
? = 37°, S = 1m.
Frictional force f = ?R
mg sin ? = ? R ..(1)
mg cos ?? ..(2)
so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J
10. a =
) m M ( 2
F
?
(given)
a) from fig (1)
ma = ?
k
R
1
and R
1
= mg
? ? =
1
R
ma
=
g ) m M ( 2
F
?
b) Frictional force acting on the smaller block f = ?R =
) m M ( 2
F m
mg
g ) m M ( 2
F
?
?
? ?
?
c) Work done w = fs s = d
w = d
) m M ( 2
mF
?
?
=
) m M ( 2
mFd
?
?
11. Weight = 2000 N, S = 20m, ? = 0.2
a) R + Psin ? - 2000 = 0 ..(1)
P cos ? - 0.2 R =0 ..(2)
From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0
P =
? ? ? sin 2 . 0 cos
400
..(3)
So, work done by the person, W = PS cos ? =
? ? ?
?
sin 2 . 0 cos
cos 8000
=
? ? sin 2 . 0 1
8000
=
? ? tan 5
40000
b) For minimum magnitude of force from equn(1)
d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2
putting the value in equn (3)
W =
? ? tan 5
40000
=
) 2 . 5 (
40000
= 7690 J
12. w = 100 N,  ? = 37°,  s = 2m
R ?
m b g ?
m b a ? F ?
R ?
?R ?
1 m ?
mg ?
37° ?
M
?
F ?
m
?
R 1 ?
? k R 1 ?
ma ?
mg ?
?R 1 ?
R 2 ?
f
?
ma ?
mg ?
?R 2 ?
? ?
R
?
P
?
0.2R ?
2000 N ?
Chapter 8
8.3
Force F= mg sin 37° = 100 × 0.60 = 60 N
So, work done, when the force is parallel to incline.
w = Fs cos ? = 60 × 2 × cos ? = 120 J
In ?ABC AB= 2m
CB = 37°
so, h = C = 1m
?work done when the force in horizontal direction
W = mgh = 100 × 1.2 = 120 J ?
13. m = 500 kg, s = 25m, u = 72km/h= 20 m/s, v = 0
(-a) =
S 2
u v
2 2
?
? a =
50
400
= 8m/sec
2
Frictional force f = ma = 500 × 8 = 4000 N
14. m = 500 kg, u = 0, v = 72 km/h = 20m/s
a =
s 2
u v
2 2
?
=
50
400
= 8m/sec
2
force needed to accelerate the car F = ma = 500 × 8 = 4000 N
15. Given, v = a x (uniformly accelerated motion)
displacement s = d – 0 = d
putting x = 0, v
1
= 0
putting x = d, v
2
= a d
a =
s 2
u v
2
2
2
2
?
=
d 2
d a
2
=
2
a
2
force f = ma =
2
ma
2
work done w = FS cos ? = d
2
ma
2
? =
2
d ma
2
?
16. a) m = 2kg, ? = 37°, F = 20 N
From the free body diagram
F = (2g sin ?) + ma ? a = (20 – 20 sin ?)/s = 4m/sec
2
S = ut + ½ at
2
(u = 0, t = 1s, a = 1.66)
= 2m
So, work, done w = Fs = 20 × 2 = 40 J
b) If W = 40 J
S =
F
W
=
20
40
h = 2 sin 37° = 1.2 m
So, work done W = –mgh = – 20 × 1.2 = –24 J
c) v = u + at  = 4 × 10 = 40 m/sec
So, K.E.  = ½ mv
2
= ½ × 2 × 16 = 16 J
17. m = 2kg, ? = 37°, F = 20 N, a = 10 m/sec
2
a) t = 1sec
So, s= ut + ½ at
2
= 5m
37°
?
A ?
A ?
A ?
B ?
v=0
?
v=20 m/s
m=500 kg ?
–a
?
25m
?
a
?
R
mg
?
ma f
?
500 kg ?
a
?
25m
?
R
mg
?
F
F
?
ma
?
ma ?
2g cos ? ?
20N
?
R
?
ma 2gsin ? ?
20N ?
ma ?
mg cos ? ?
20N
?
R
?
mg sin ? ?
?R ?
h
?
37°
?
5m ?
C ?
A ?
B ?
37°
?
Page 4

8.1
SOLUTIONS TO CONCEPTS
CHAPTER – 8
1. M = m
c
+ m
b
= 90kg
u =  6 km/h = 1.666 m/sec
v = 12 km/h = 3.333 m/sec
Increase in K.E. = ½ Mv
2
– ½ Mu
2
= ½ 90 × (3.333)
2
– ½ × 90 × (1.66)
2
= 494.5 – 124.6 = 374.8 ? 375 J
2. m
b
= 2 kg.
u = 10 m/sec
a = 3 m/aec
2
t = 5 sec
v = u + at = 10 + 3 I 5 = 25 m/sec.
?F.K.E = ½ mv
2
= ½ × 2 × 625 = 625 J.
3. F = 100 N
S = 4m, ? = 0°
? = S . F
? ?
???100 × 4 = 400 J ?
4. m = 5 kg
? = 30°
S = 10 m
F = mg
So, work done by the force of gravity
? = mgh = 5 × 9.8 × 5 = 245 J ?
5. F= 2.50N, S = 2.5m,  m =15g = 0.015kg.
So, w = F × S ? a =
m
F
=
015 . 0
5 . 2
=
3
500
m/s
2
=F × S cos 0° (acting along the same line)
= 2.5 × 2.5 = 6.25J
Let the velocity of the body at b = U. Applying work-energy principle ½ mv
2
– 0 = 6.25
? V =
015 . 0
2 25 . 6 ?
= 28.86 m/sec.
So, time taken to travel from A to B.
? t =
a
u v ?
=
500
3 86 . 28 ?
? Average power =
t
W
=
3 ) 86 . 28 (
500 25 . 6
?
?
= 36.1
6. Given
j
ˆ
3 i
ˆ
2 r
1
? ?
?
j
ˆ
2 i
ˆ
3 r
2
? ?
So, displacement vector is given by,
2 1
r r r
? ? ?
? ? ? j
ˆ
i
ˆ
) j
ˆ
3 i
ˆ
2 ( ) j
ˆ
2 i
ˆ
3 ( r ? ? ? ? ? ?
?
u=1.66 m/s ?
90kg ? 90kg ?
v=3.33 m/s ?
u=10 m/s ?
2 kg ? a
?
= 3m/s
2
F ?
4m ?
R ?
100 N ?
mg ?
30°
?
5 ?
mg ?
F ?
5 log ?
10m ?
30°
?
v ? ?
A ? B ?
Chapter 8
8.2
So, work done = s F
?
?
? = 5 × 1 + 5(-1) = 0
7. m
b
= 2kg, s = 40m, a = 0.5m/sec
2
So, force applied by the man on the box
F = m
b
a = 2 × (0.5) = 1 N
? = FS = 1 × 40 = 40 J ?
8. Given that F= a + bx
Where a and b are constants.
So, work done by this force during this force during the displacement x = 0 and  x = d is given
by
W =
? ?
? ?
d
0
d
0
dx ) bx a ( dx F = ax + (bx
2
/2) = [a + ½ bd] d
9. m
b
= 250g = .250 kg
? = 37°, S = 1m.
Frictional force f = ?R
mg sin ? = ? R ..(1)
mg cos ?? ..(2)
so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J
10. a =
) m M ( 2
F
?
(given)
a) from fig (1)
ma = ?
k
R
1
and R
1
= mg
? ? =
1
R
ma
=
g ) m M ( 2
F
?
b) Frictional force acting on the smaller block f = ?R =
) m M ( 2
F m
mg
g ) m M ( 2
F
?
?
? ?
?
c) Work done w = fs s = d
w = d
) m M ( 2
mF
?
?
=
) m M ( 2
mFd
?
?
11. Weight = 2000 N, S = 20m, ? = 0.2
a) R + Psin ? - 2000 = 0 ..(1)
P cos ? - 0.2 R =0 ..(2)
From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0
P =
? ? ? sin 2 . 0 cos
400
..(3)
So, work done by the person, W = PS cos ? =
? ? ?
?
sin 2 . 0 cos
cos 8000
=
? ? sin 2 . 0 1
8000
=
? ? tan 5
40000
b) For minimum magnitude of force from equn(1)
d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2
putting the value in equn (3)
W =
? ? tan 5
40000
=
) 2 . 5 (
40000
= 7690 J
12. w = 100 N,  ? = 37°,  s = 2m
R ?
m b g ?
m b a ? F ?
R ?
?R ?
1 m ?
mg ?
37° ?
M
?
F ?
m
?
R 1 ?
? k R 1 ?
ma ?
mg ?
?R 1 ?
R 2 ?
f
?
ma ?
mg ?
?R 2 ?
? ?
R
?
P
?
0.2R ?
2000 N ?
Chapter 8
8.3
Force F= mg sin 37° = 100 × 0.60 = 60 N
So, work done, when the force is parallel to incline.
w = Fs cos ? = 60 × 2 × cos ? = 120 J
In ?ABC AB= 2m
CB = 37°
so, h = C = 1m
?work done when the force in horizontal direction
W = mgh = 100 × 1.2 = 120 J ?
13. m = 500 kg, s = 25m, u = 72km/h= 20 m/s, v = 0
(-a) =
S 2
u v
2 2
?
? a =
50
400
= 8m/sec
2
Frictional force f = ma = 500 × 8 = 4000 N
14. m = 500 kg, u = 0, v = 72 km/h = 20m/s
a =
s 2
u v
2 2
?
=
50
400
= 8m/sec
2
force needed to accelerate the car F = ma = 500 × 8 = 4000 N
15. Given, v = a x (uniformly accelerated motion)
displacement s = d – 0 = d
putting x = 0, v
1
= 0
putting x = d, v
2
= a d
a =
s 2
u v
2
2
2
2
?
=
d 2
d a
2
=
2
a
2
force f = ma =
2
ma
2
work done w = FS cos ? = d
2
ma
2
? =
2
d ma
2
?
16. a) m = 2kg, ? = 37°, F = 20 N
From the free body diagram
F = (2g sin ?) + ma ? a = (20 – 20 sin ?)/s = 4m/sec
2
S = ut + ½ at
2
(u = 0, t = 1s, a = 1.66)
= 2m
So, work, done w = Fs = 20 × 2 = 40 J
b) If W = 40 J
S =
F
W
=
20
40
h = 2 sin 37° = 1.2 m
So, work done W = –mgh = – 20 × 1.2 = –24 J
c) v = u + at  = 4 × 10 = 40 m/sec
So, K.E.  = ½ mv
2
= ½ × 2 × 16 = 16 J
17. m = 2kg, ? = 37°, F = 20 N, a = 10 m/sec
2
a) t = 1sec
So, s= ut + ½ at
2
= 5m
37°
?
A ?
A ?
A ?
B ?
v=0
?
v=20 m/s
m=500 kg ?
–a
?
25m
?
a
?
R
mg
?
ma f
?
500 kg ?
a
?
25m
?
R
mg
?
F
F
?
ma
?
ma ?
2g cos ? ?
20N
?
R
?
ma 2gsin ? ?
20N ?
ma ?
mg cos ? ?
20N
?
R
?
mg sin ? ?
?R ?
h
?
37°
?
5m ?
C ?
A ?
B ?
37°
?
Chapter 8
8.4
Work done by the applied force w = FS cos 0° = 20 × 5 = 100 J
b) BC (h) = 5 sin 37° = 3m
So, work done by the weight W = mgh = 2 × 10 × 3 = 60 J
c) So, frictional force f = mg sin ?
work done by the frictional forces w = fs cos0° = (mg sin ?) s = 20 × 0.60 × 5  = 60 J ?
18. Given, m = 25o g = 0.250kg,
u = 40 cm/sec = 0.4m/sec
? = 0.1, v=0
Here, ? R = ma {where, a = deceleration}
a =
m
R ?
=
m
mg ?
= ?g = 0.1 × 9.8 = 0.98 m/sec
2
S =
a 2
u v
2 2
?
= 0.082m = 8.2 cm
Again, work done against friction is given by
– w = ? RS cos ?
= 0.1 × 2.5 × 0.082 × 1 ( ? = 0°) = 0.02 J
? W = – 0.02 J
?
19. h = 50m, m = 1.8 × 10
5
kg/hr, P = 100 watt,
P.E. = mgh = 1.8 × 10
5
× 9.8 × 50 = 882 × 10
5
J/hr
Because, half the potential energy is converted into electricity,
Electrical energy ½ P.E. = 441 × 10
5
J/hr
So, power in watt (J/sec) is given by =
3600
10 441
5
?
? number of 100 W lamps, that can be lit
100 3600
10 441
5
?
?
= 122.5 ?122
20. m = 6kg, h = 2m
P.E. at a height ‘2m’ = mgh = 6 × (9.8) × 2 = 117.6 J
P.E. at floor = 0
Loss in P.E. = 117.6 – 0 = 117. 6 J ? 118 J
21. h = 40m, u = 50 m/sec
Let the speed be ‘v’ when it strikes the ground.
Applying law of conservation of energy
mgh + ½ mu
2
= ½ mv
2
? 10 × 40 + (1/2) × 2500 = ½ v
2
? v
2
= 3300 ? v = 57.4 m/sec ?58 m/sec
22. t = 1 min 57.56 sec = 11.56 sec, p= 400 W, s =200 m
p =
t
w
, Work w = pt = 460 × 117.56 J
Again, W = FS =
200
56 . 117 460 ?
= 270.3 N ? 270 N
23. S = 100 m, t = 10.54 sec, m = 50 kg
The motion can be assumed to be uniform because the time taken for acceleration is
minimum.
Page 5

8.1
SOLUTIONS TO CONCEPTS
CHAPTER – 8
1. M = m
c
+ m
b
= 90kg
u =  6 km/h = 1.666 m/sec
v = 12 km/h = 3.333 m/sec
Increase in K.E. = ½ Mv
2
– ½ Mu
2
= ½ 90 × (3.333)
2
– ½ × 90 × (1.66)
2
= 494.5 – 124.6 = 374.8 ? 375 J
2. m
b
= 2 kg.
u = 10 m/sec
a = 3 m/aec
2
t = 5 sec
v = u + at = 10 + 3 I 5 = 25 m/sec.
?F.K.E = ½ mv
2
= ½ × 2 × 625 = 625 J.
3. F = 100 N
S = 4m, ? = 0°
? = S . F
? ?
???100 × 4 = 400 J ?
4. m = 5 kg
? = 30°
S = 10 m
F = mg
So, work done by the force of gravity
? = mgh = 5 × 9.8 × 5 = 245 J ?
5. F= 2.50N, S = 2.5m,  m =15g = 0.015kg.
So, w = F × S ? a =
m
F
=
015 . 0
5 . 2
=
3
500
m/s
2
=F × S cos 0° (acting along the same line)
= 2.5 × 2.5 = 6.25J
Let the velocity of the body at b = U. Applying work-energy principle ½ mv
2
– 0 = 6.25
? V =
015 . 0
2 25 . 6 ?
= 28.86 m/sec.
So, time taken to travel from A to B.
? t =
a
u v ?
=
500
3 86 . 28 ?
? Average power =
t
W
=
3 ) 86 . 28 (
500 25 . 6
?
?
= 36.1
6. Given
j
ˆ
3 i
ˆ
2 r
1
? ?
?
j
ˆ
2 i
ˆ
3 r
2
? ?
So, displacement vector is given by,
2 1
r r r
? ? ?
? ? ? j
ˆ
i
ˆ
) j
ˆ
3 i
ˆ
2 ( ) j
ˆ
2 i
ˆ
3 ( r ? ? ? ? ? ?
?
u=1.66 m/s ?
90kg ? 90kg ?
v=3.33 m/s ?
u=10 m/s ?
2 kg ? a
?
= 3m/s
2
F ?
4m ?
R ?
100 N ?
mg ?
30°
?
5 ?
mg ?
F ?
5 log ?
10m ?
30°
?
v ? ?
A ? B ?
Chapter 8
8.2
So, work done = s F
?
?
? = 5 × 1 + 5(-1) = 0
7. m
b
= 2kg, s = 40m, a = 0.5m/sec
2
So, force applied by the man on the box
F = m
b
a = 2 × (0.5) = 1 N
? = FS = 1 × 40 = 40 J ?
8. Given that F= a + bx
Where a and b are constants.
So, work done by this force during this force during the displacement x = 0 and  x = d is given
by
W =
? ?
? ?
d
0
d
0
dx ) bx a ( dx F = ax + (bx
2
/2) = [a + ½ bd] d
9. m
b
= 250g = .250 kg
? = 37°, S = 1m.
Frictional force f = ?R
mg sin ? = ? R ..(1)
mg cos ?? ..(2)
so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J
10. a =
) m M ( 2
F
?
(given)
a) from fig (1)
ma = ?
k
R
1
and R
1
= mg
? ? =
1
R
ma
=
g ) m M ( 2
F
?
b) Frictional force acting on the smaller block f = ?R =
) m M ( 2
F m
mg
g ) m M ( 2
F
?
?
? ?
?
c) Work done w = fs s = d
w = d
) m M ( 2
mF
?
?
=
) m M ( 2
mFd
?
?
11. Weight = 2000 N, S = 20m, ? = 0.2
a) R + Psin ? - 2000 = 0 ..(1)
P cos ? - 0.2 R =0 ..(2)
From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0
P =
? ? ? sin 2 . 0 cos
400
..(3)
So, work done by the person, W = PS cos ? =
? ? ?
?
sin 2 . 0 cos
cos 8000
=
? ? sin 2 . 0 1
8000
=
? ? tan 5
40000
b) For minimum magnitude of force from equn(1)
d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2
putting the value in equn (3)
W =
? ? tan 5
40000
=
) 2 . 5 (
40000
= 7690 J
12. w = 100 N,  ? = 37°,  s = 2m
R ?
m b g ?
m b a ? F ?
R ?
?R ?
1 m ?
mg ?
37° ?
M
?
F ?
m
?
R 1 ?
? k R 1 ?
ma ?
mg ?
?R 1 ?
R 2 ?
f
?
ma ?
mg ?
?R 2 ?
? ?
R
?
P
?
0.2R ?
2000 N ?
Chapter 8
8.3
Force F= mg sin 37° = 100 × 0.60 = 60 N
So, work done, when the force is parallel to incline.
w = Fs cos ? = 60 × 2 × cos ? = 120 J
In ?ABC AB= 2m
CB = 37°
so, h = C = 1m
?work done when the force in horizontal direction
W = mgh = 100 × 1.2 = 120 J ?
13. m = 500 kg, s = 25m, u = 72km/h= 20 m/s, v = 0
(-a) =
S 2
u v
2 2
?
? a =
50
400
= 8m/sec
2
Frictional force f = ma = 500 × 8 = 4000 N
14. m = 500 kg, u = 0, v = 72 km/h = 20m/s
a =
s 2
u v
2 2
?
=
50
400
= 8m/sec
2
force needed to accelerate the car F = ma = 500 × 8 = 4000 N
15. Given, v = a x (uniformly accelerated motion)
displacement s = d – 0 = d
putting x = 0, v
1
= 0
putting x = d, v
2
= a d
a =
s 2
u v
2
2
2
2
?
=
d 2
d a
2
=
2
a
2
force f = ma =
2
ma
2
work done w = FS cos ? = d
2
ma
2
? =
2
d ma
2
?
16. a) m = 2kg, ? = 37°, F = 20 N
From the free body diagram
F = (2g sin ?) + ma ? a = (20 – 20 sin ?)/s = 4m/sec
2
S = ut + ½ at
2
(u = 0, t = 1s, a = 1.66)
= 2m
So, work, done w = Fs = 20 × 2 = 40 J
b) If W = 40 J
S =
F
W
=
20
40
h = 2 sin 37° = 1.2 m
So, work done W = –mgh = – 20 × 1.2 = –24 J
c) v = u + at  = 4 × 10 = 40 m/sec
So, K.E.  = ½ mv
2
= ½ × 2 × 16 = 16 J
17. m = 2kg, ? = 37°, F = 20 N, a = 10 m/sec
2
a) t = 1sec
So, s= ut + ½ at
2
= 5m
37°
?
A ?
A ?
A ?
B ?
v=0
?
v=20 m/s
m=500 kg ?
–a
?
25m
?
a
?
R
mg
?
ma f
?
500 kg ?
a
?
25m
?
R
mg
?
F
F
?
ma
?
ma ?
2g cos ? ?
20N
?
R
?
ma 2gsin ? ?
20N ?
ma ?
mg cos ? ?
20N
?
R
?
mg sin ? ?
?R ?
h
?
37°
?
5m ?
C ?
A ?
B ?
37°
?
Chapter 8
8.4
Work done by the applied force w = FS cos 0° = 20 × 5 = 100 J
b) BC (h) = 5 sin 37° = 3m
So, work done by the weight W = mgh = 2 × 10 × 3 = 60 J
c) So, frictional force f = mg sin ?
work done by the frictional forces w = fs cos0° = (mg sin ?) s = 20 × 0.60 × 5  = 60 J ?
18. Given, m = 25o g = 0.250kg,
u = 40 cm/sec = 0.4m/sec
? = 0.1, v=0
Here, ? R = ma {where, a = deceleration}
a =
m
R ?
=
m
mg ?
= ?g = 0.1 × 9.8 = 0.98 m/sec
2
S =
a 2
u v
2 2
?
= 0.082m = 8.2 cm
Again, work done against friction is given by
– w = ? RS cos ?
= 0.1 × 2.5 × 0.082 × 1 ( ? = 0°) = 0.02 J
? W = – 0.02 J
?
19. h = 50m, m = 1.8 × 10
5
kg/hr, P = 100 watt,
P.E. = mgh = 1.8 × 10
5
× 9.8 × 50 = 882 × 10
5
J/hr
Because, half the potential energy is converted into electricity,
Electrical energy ½ P.E. = 441 × 10
5
J/hr
So, power in watt (J/sec) is given by =
3600
10 441
5
?
? number of 100 W lamps, that can be lit
100 3600
10 441
5
?
?
= 122.5 ?122
20. m = 6kg, h = 2m
P.E. at a height ‘2m’ = mgh = 6 × (9.8) × 2 = 117.6 J
P.E. at floor = 0
Loss in P.E. = 117.6 – 0 = 117. 6 J ? 118 J
21. h = 40m, u = 50 m/sec
Let the speed be ‘v’ when it strikes the ground.
Applying law of conservation of energy
mgh + ½ mu
2
= ½ mv
2
? 10 × 40 + (1/2) × 2500 = ½ v
2
? v
2
= 3300 ? v = 57.4 m/sec ?58 m/sec
22. t = 1 min 57.56 sec = 11.56 sec, p= 400 W, s =200 m
p =
t
w
, Work w = pt = 460 × 117.56 J
Again, W = FS =
200
56 . 117 460 ?
= 270.3 N ? 270 N
23. S = 100 m, t = 10.54 sec, m = 50 kg
The motion can be assumed to be uniform because the time taken for acceleration is
minimum.
Chapter 8
8.5
a) Speed v = S/t = 9.487 e/s
So, K.E. = ½ mv
2
= 2250 J
b) Weight = mg = 490 J
given R = mg /10 = 49 J
so, work done against resistance W
F
= – RS = – 49 × 100 = – 4900 J
c) To maintain her uniform speed, she has to exert 4900 j of energy to over come friction
P =
t
W
= 4900 / 10.54 = 465 W
24. h = 10 m
flow rate = (m/t) = 30 kg/min = 0.5 kg/sec
power P =
t
mgh
= (0.5) × 9.8  × 10 = 49 W
So, horse power (h.p) P/746 = 49/746 = 6.6 × 10
–2
hp
25. m = 200g = 0.2kg, h = 150cm = 1.5m, v = 3m/sec, t = 1 sec
Total work done = ½ mv
2
+ mgh = (1/2) × (0.2) ×9 + (0.2) × (9.8) × (1.5) = 3.84 J
h.p. used =
746
84 . 3
= 5.14 × 10
–3
26. m = 200 kg, s = 12m, t = 1 min = 60 sec
So, work  W = F cos ? = mgs cos0° [ ? = 0°, for minimum work]
= 2000 × 10 × 12 = 240000 J
So, power p =
t
W
=
60
240000
= 4000 watt
h.p =
746
4000
= 5.3 hp. ?
27. The specification given by the company are
U = 0, m = 95 kg, P
m
= 3.5 hp
V
m
= 60 km/h = 50/3 m/sec t
m
= 5 sec
So, the maximum acceleration that can be produced is given by,
a =
5
0 ) 3 / 50 ( ?
=
3
10
So, the driving force is given by
F = ma = 95 ×
3
10
=
3
950
N
So, the velocity that can be attained by maximum h.p. white supplying
3
950
will be
v =
F
p
? v =
950
5 746 5 . 3 ? ?
= 8.2 m/sec.
Because, the scooter can reach a maximum of 8.s m/sec while producing a force of 950/3 N,
the specifications given are some what over claimed.
28. Given m = 30kg, v = 40 cm/sec = 0.4 m/sec s = 2m
From the free body diagram, the force given by the chain is,
F = (ma – mg) = m(a – g) [where a = acceleration of the block]
a =
2s
u2)  (v2
=
4 . 0
16 . 0
= 0.04 m/sec
2
mg
?
F
?
ma
?
mg
?
F
?
```
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