Work, Energy and Power-1 Practice Questions - DPP for NEET

 Page 1


(1) (a)  Here
ˆ ˆˆ
F i 2 j 3k =-++
r
 &
 d
r
 = (0 – 0) 
ˆ
i + (0 – 0)
ˆ
j + (4 – 0)
ˆ
k = 4
ˆ
k
\ W (W ork done)  = F.d
r r
 = 
ˆ ˆˆ
( i 2j 3k) -++ . 4
ˆ
k = 12 J
(2) (a) The minimum force with a body is to be pulled up along
the inclined plane is mg (sin q + m cos q)
Work done, 
W F.d =
r r
=  Fd cos q = mg (sin q + m cos q)  × d
= 5 × 9.8 (sin 60º + 0.2 cos 60º) × 2 = 98.08 J
(3) (d) W = 
5
0
F dx
ò
= 
5
2
0
(7 2x 3x ) dx -+
ò
= [ ]
0
5
x 7 – 
5
2
0
2x
2
éù
êú
êú
ëû
 + 
5
3
0
3x
3
éù
êú
êú
ëû
 = 135 Joule
(4) (d) Given that, power =  Fv = P = constant
or    m 
dv
dt
 v = P [as F = ma = 
mdv
dt
]
or v dv
ò
   = 
P
dt
m
ò Þ
 
2
v
2
  = 
P
m
t + C
1
Now as initially , the body is at rest
i.e v = 0 at t = 0 so, C
1
 = 0
\  v  =
2Pt
m
(5) (b)  By definition v = 
ds
dt
or 
ds
dt
 = 
1/2
2Pt
m
æö
ç÷
èø
Þ ds
ò
 = 
1/2
2Pt
dt
m
æö
ç÷
èø
ò
 Þ s = 
1/2
2P
m
æö
ç÷
èø
2
3
t
3/2
 + C
2
Now as t = 0, s = 0, so C
2 
= 0
       s = 
1/2
8P
9m
æö
ç÷
èø
 t
3/2
(6) (c) The force acting on the particle = 
mdv
dt
Power of the force = 
mdv
dt
æö
ç÷
èø
 v = k (constant)
Þ  m 
2
v
2
 = kt + c .....(1)
At  t = 0, v = u \  c = 
2
mu
2
Now from (1),  m 
2
v
2
 = kt + 
2
mu
2
Þ  
1
2
m (v
2 
– u
2
) = kt .....(2)
Again  
mdv
dt
 v = k
Þ  m.v 
dx
dv
 v = k  Þ  mv
2
 dv = kdx
Intergrating,
1
3
m (v
3 
– u
3
) = kx .....(3)
From (2) and (3),
t  = 
3
2
 
22
33
vu
vu
æö
-
ç÷
- èø
 (x)
(7) (b) Mass of the chain hanging  =  4 × 3 = 12 kg
Shift in center of gravity   = 4/2 = 2m
Work done,  W = mgh = 12 × 9.8 × 2 = 235 .2 J
(8) (b)  Mass of 2 litre, water  = 2 kg
Total mass to be lifted  = 2 + 0.5 = 2.5 kg
Work done ,  W  = mgh = 2.5 × 9.8 × 6 = 147 J
(9) (b) The following two forces are acting on the body
(i) Weight mg is acting vertically downward
(ii) The push of the air is acting upward.
As the body is accelerating downward, the resultant force
is (mg – F)
Workdone by  the resultant force to fall through a vertical
distance of  20 m = (mg – F) × 20 joule
Gain in the kinetic energy = 
1
2
mv
2
 
Now the workdone by the resultant force is equal to the
change in kinetic energy i.e.
(mg – F) 20 = 
1
2
mv
2 
(From work-energy theorem)
a
F
mg
Page 2


(1) (a)  Here
ˆ ˆˆ
F i 2 j 3k =-++
r
 &
 d
r
 = (0 – 0) 
ˆ
i + (0 – 0)
ˆ
j + (4 – 0)
ˆ
k = 4
ˆ
k
\ W (W ork done)  = F.d
r r
 = 
ˆ ˆˆ
( i 2j 3k) -++ . 4
ˆ
k = 12 J
(2) (a) The minimum force with a body is to be pulled up along
the inclined plane is mg (sin q + m cos q)
Work done, 
W F.d =
r r
=  Fd cos q = mg (sin q + m cos q)  × d
= 5 × 9.8 (sin 60º + 0.2 cos 60º) × 2 = 98.08 J
(3) (d) W = 
5
0
F dx
ò
= 
5
2
0
(7 2x 3x ) dx -+
ò
= [ ]
0
5
x 7 – 
5
2
0
2x
2
éù
êú
êú
ëû
 + 
5
3
0
3x
3
éù
êú
êú
ëû
 = 135 Joule
(4) (d) Given that, power =  Fv = P = constant
or    m 
dv
dt
 v = P [as F = ma = 
mdv
dt
]
or v dv
ò
   = 
P
dt
m
ò Þ
 
2
v
2
  = 
P
m
t + C
1
Now as initially , the body is at rest
i.e v = 0 at t = 0 so, C
1
 = 0
\  v  =
2Pt
m
(5) (b)  By definition v = 
ds
dt
or 
ds
dt
 = 
1/2
2Pt
m
æö
ç÷
èø
Þ ds
ò
 = 
1/2
2Pt
dt
m
æö
ç÷
èø
ò
 Þ s = 
1/2
2P
m
æö
ç÷
èø
2
3
t
3/2
 + C
2
Now as t = 0, s = 0, so C
2 
= 0
       s = 
1/2
8P
9m
æö
ç÷
èø
 t
3/2
(6) (c) The force acting on the particle = 
mdv
dt
Power of the force = 
mdv
dt
æö
ç÷
èø
 v = k (constant)
Þ  m 
2
v
2
 = kt + c .....(1)
At  t = 0, v = u \  c = 
2
mu
2
Now from (1),  m 
2
v
2
 = kt + 
2
mu
2
Þ  
1
2
m (v
2 
– u
2
) = kt .....(2)
Again  
mdv
dt
 v = k
Þ  m.v 
dx
dv
 v = k  Þ  mv
2
 dv = kdx
Intergrating,
1
3
m (v
3 
– u
3
) = kx .....(3)
From (2) and (3),
t  = 
3
2
 
22
33
vu
vu
æö
-
ç÷
- èø
 (x)
(7) (b) Mass of the chain hanging  =  4 × 3 = 12 kg
Shift in center of gravity   = 4/2 = 2m
Work done,  W = mgh = 12 × 9.8 × 2 = 235 .2 J
(8) (b)  Mass of 2 litre, water  = 2 kg
Total mass to be lifted  = 2 + 0.5 = 2.5 kg
Work done ,  W  = mgh = 2.5 × 9.8 × 6 = 147 J
(9) (b) The following two forces are acting on the body
(i) Weight mg is acting vertically downward
(ii) The push of the air is acting upward.
As the body is accelerating downward, the resultant force
is (mg – F)
Workdone by  the resultant force to fall through a vertical
distance of  20 m = (mg – F) × 20 joule
Gain in the kinetic energy = 
1
2
mv
2
 
Now the workdone by the resultant force is equal to the
change in kinetic energy i.e.
(mg – F) 20 = 
1
2
mv
2 
(From work-energy theorem)
a
F
mg
DPP/ P 12
35
   or (50 – F) 20 =
1
2
× 5 × (10)
2
or 50 – F = 12.5 or  F = 50 – 12.5
\ F = 37.5 N
Work done by the force = – 37.5 × 20 = – 750 joule
(The negative sign is used because the push of the air is
upwards while the displacement is downwards.)
(10) (a)
F cos 45°
F sin 45°
R
mg
mR
The different forces acting on the block are shown in fig.
Now we have
R + F sin 45° = m g ............(1)
F cos 45° = m R ............(2)
From equation (1) and (2)
\  
mg
F
cos45º sin 45º
m
=
+m
Substituting the given values, we have
     
0.20 (5 9.78)
F
(0.707) (0.20 0.707)
´´
=
+´
 = 11.55 N
The block is pulled through a horizontal distance
  r = 20 metre
Hence, the work done
W = F cos 45° × r  = (11. 55 × 0.707) × 20 = 163. 32 Joule
(11) (c)
l/5
Mass of the hanging part of the chain = (m/5) The weight
mg/5 acts at the centre of gravity of the hanging chain, i.e.,
at a distance = l/10 below the surface of a table.
The gain in potential energy in pulling the hanging part on
the table.
U = 
mg
5
 × 
10
l
 = 
mg
10
l
\  Work done = U = mg l/50
(12) (a) At maximum speed all the power is used to overcome
the resistance to motion. Hence if the maximum speed is v ,
then 50000 = 1000 × v or v = 50 m/s
At 25 m/s, let the pull of the engine be P , then the power
or  P = 
50,000
25
 = 2000 N
Now resultant force = 2000 – 1000 = 1000 N
Applying Newton's law ; F = ma, we have
     1000 = 1000 a or a = 1.0 m/s
2
(13) (a) 1 mole i.e.235 gm of uranium contains 6 × 10
23
 atoms, so
2 kg i.e. 2 × 10
3
 gm of uranium will contain
= 
3 23
2 10 6 10
235
´ ´´
 atoms = 5.106 × 10
24
 atoms
Now as in each fission only one uranium atom is con-
sumed i.e. Energy yield per uranium atom
= 185 MeV = 185 × 1.6 × 10 
–13
 J = 2.96 × 10
–11
 J
So Energy produced by 2 kg uranium
= (No. of atoms ) × (energy /atom)
= 5.106 × 10
24
 × 2.96 × 10
–11
 = 1.514 × 10
–14
 J
As 2 kg uranium is consumed in 30 days i.e. 1.51 × 10
–14
 J
of energy is produced in the reactor in 30 days i.e.
2.592 × 10
6
 sec
So, power output of reactor
= 
E
t
 = 
14
6
1.514 10 J
2.592 10 S
´
´
 = 58.4 MW
(14) (c) When the vehicle of mass m is moving with velocity v,
the kinetic energy of the where K = 
1
2
 mv
2 
and if S is the
stopping distance, work done by the friction
W = FS cos q = m MgS cos 180º = – m MgS
So by Work-Energy theorem,
W = D K = K
f
 – k
i
Þ
 
– m MgS = 0 – 
1
2
 Mv
2
  Þ  S = 
2
v
2g m
(15) (a) As T = (2p/w),
so w = 2p/(3.15 × 10
7
) = 1.99 × 10
–7
 rad/s
Now  v = rw = 1.5 × 10
11
 × 1.99 × 10
–7
  » 3 × 10
4
 m/s
Now by work - energy theorem ,
W = K
f
 – K
i 
 = 0 – 
1
2
 mv
2
    
= – 
1
2
 × 6 × 10
24
 (3 × 10
4
)
2  
= – 2.7 × 10
33 
J
Negative sign means force is opposite to the motion.
(16) (b) As the particle is moving in a circle, so
2
2
mvk
r
r
=
 Now K.E = 
1
2
 mv
2 
 = 
k
2r
Page 3


(1) (a)  Here
ˆ ˆˆ
F i 2 j 3k =-++
r
 &
 d
r
 = (0 – 0) 
ˆ
i + (0 – 0)
ˆ
j + (4 – 0)
ˆ
k = 4
ˆ
k
\ W (W ork done)  = F.d
r r
 = 
ˆ ˆˆ
( i 2j 3k) -++ . 4
ˆ
k = 12 J
(2) (a) The minimum force with a body is to be pulled up along
the inclined plane is mg (sin q + m cos q)
Work done, 
W F.d =
r r
=  Fd cos q = mg (sin q + m cos q)  × d
= 5 × 9.8 (sin 60º + 0.2 cos 60º) × 2 = 98.08 J
(3) (d) W = 
5
0
F dx
ò
= 
5
2
0
(7 2x 3x ) dx -+
ò
= [ ]
0
5
x 7 – 
5
2
0
2x
2
éù
êú
êú
ëû
 + 
5
3
0
3x
3
éù
êú
êú
ëû
 = 135 Joule
(4) (d) Given that, power =  Fv = P = constant
or    m 
dv
dt
 v = P [as F = ma = 
mdv
dt
]
or v dv
ò
   = 
P
dt
m
ò Þ
 
2
v
2
  = 
P
m
t + C
1
Now as initially , the body is at rest
i.e v = 0 at t = 0 so, C
1
 = 0
\  v  =
2Pt
m
(5) (b)  By definition v = 
ds
dt
or 
ds
dt
 = 
1/2
2Pt
m
æö
ç÷
èø
Þ ds
ò
 = 
1/2
2Pt
dt
m
æö
ç÷
èø
ò
 Þ s = 
1/2
2P
m
æö
ç÷
èø
2
3
t
3/2
 + C
2
Now as t = 0, s = 0, so C
2 
= 0
       s = 
1/2
8P
9m
æö
ç÷
èø
 t
3/2
(6) (c) The force acting on the particle = 
mdv
dt
Power of the force = 
mdv
dt
æö
ç÷
èø
 v = k (constant)
Þ  m 
2
v
2
 = kt + c .....(1)
At  t = 0, v = u \  c = 
2
mu
2
Now from (1),  m 
2
v
2
 = kt + 
2
mu
2
Þ  
1
2
m (v
2 
– u
2
) = kt .....(2)
Again  
mdv
dt
 v = k
Þ  m.v 
dx
dv
 v = k  Þ  mv
2
 dv = kdx
Intergrating,
1
3
m (v
3 
– u
3
) = kx .....(3)
From (2) and (3),
t  = 
3
2
 
22
33
vu
vu
æö
-
ç÷
- èø
 (x)
(7) (b) Mass of the chain hanging  =  4 × 3 = 12 kg
Shift in center of gravity   = 4/2 = 2m
Work done,  W = mgh = 12 × 9.8 × 2 = 235 .2 J
(8) (b)  Mass of 2 litre, water  = 2 kg
Total mass to be lifted  = 2 + 0.5 = 2.5 kg
Work done ,  W  = mgh = 2.5 × 9.8 × 6 = 147 J
(9) (b) The following two forces are acting on the body
(i) Weight mg is acting vertically downward
(ii) The push of the air is acting upward.
As the body is accelerating downward, the resultant force
is (mg – F)
Workdone by  the resultant force to fall through a vertical
distance of  20 m = (mg – F) × 20 joule
Gain in the kinetic energy = 
1
2
mv
2
 
Now the workdone by the resultant force is equal to the
change in kinetic energy i.e.
(mg – F) 20 = 
1
2
mv
2 
(From work-energy theorem)
a
F
mg
DPP/ P 12
35
   or (50 – F) 20 =
1
2
× 5 × (10)
2
or 50 – F = 12.5 or  F = 50 – 12.5
\ F = 37.5 N
Work done by the force = – 37.5 × 20 = – 750 joule
(The negative sign is used because the push of the air is
upwards while the displacement is downwards.)
(10) (a)
F cos 45°
F sin 45°
R
mg
mR
The different forces acting on the block are shown in fig.
Now we have
R + F sin 45° = m g ............(1)
F cos 45° = m R ............(2)
From equation (1) and (2)
\  
mg
F
cos45º sin 45º
m
=
+m
Substituting the given values, we have
     
0.20 (5 9.78)
F
(0.707) (0.20 0.707)
´´
=
+´
 = 11.55 N
The block is pulled through a horizontal distance
  r = 20 metre
Hence, the work done
W = F cos 45° × r  = (11. 55 × 0.707) × 20 = 163. 32 Joule
(11) (c)
l/5
Mass of the hanging part of the chain = (m/5) The weight
mg/5 acts at the centre of gravity of the hanging chain, i.e.,
at a distance = l/10 below the surface of a table.
The gain in potential energy in pulling the hanging part on
the table.
U = 
mg
5
 × 
10
l
 = 
mg
10
l
\  Work done = U = mg l/50
(12) (a) At maximum speed all the power is used to overcome
the resistance to motion. Hence if the maximum speed is v ,
then 50000 = 1000 × v or v = 50 m/s
At 25 m/s, let the pull of the engine be P , then the power
or  P = 
50,000
25
 = 2000 N
Now resultant force = 2000 – 1000 = 1000 N
Applying Newton's law ; F = ma, we have
     1000 = 1000 a or a = 1.0 m/s
2
(13) (a) 1 mole i.e.235 gm of uranium contains 6 × 10
23
 atoms, so
2 kg i.e. 2 × 10
3
 gm of uranium will contain
= 
3 23
2 10 6 10
235
´ ´´
 atoms = 5.106 × 10
24
 atoms
Now as in each fission only one uranium atom is con-
sumed i.e. Energy yield per uranium atom
= 185 MeV = 185 × 1.6 × 10 
–13
 J = 2.96 × 10
–11
 J
So Energy produced by 2 kg uranium
= (No. of atoms ) × (energy /atom)
= 5.106 × 10
24
 × 2.96 × 10
–11
 = 1.514 × 10
–14
 J
As 2 kg uranium is consumed in 30 days i.e. 1.51 × 10
–14
 J
of energy is produced in the reactor in 30 days i.e.
2.592 × 10
6
 sec
So, power output of reactor
= 
E
t
 = 
14
6
1.514 10 J
2.592 10 S
´
´
 = 58.4 MW
(14) (c) When the vehicle of mass m is moving with velocity v,
the kinetic energy of the where K = 
1
2
 mv
2 
and if S is the
stopping distance, work done by the friction
W = FS cos q = m MgS cos 180º = – m MgS
So by Work-Energy theorem,
W = D K = K
f
 – k
i
Þ
 
– m MgS = 0 – 
1
2
 Mv
2
  Þ  S = 
2
v
2g m
(15) (a) As T = (2p/w),
so w = 2p/(3.15 × 10
7
) = 1.99 × 10
–7
 rad/s
Now  v = rw = 1.5 × 10
11
 × 1.99 × 10
–7
  » 3 × 10
4
 m/s
Now by work - energy theorem ,
W = K
f
 – K
i 
 = 0 – 
1
2
 mv
2
    
= – 
1
2
 × 6 × 10
24
 (3 × 10
4
)
2  
= – 2.7 × 10
33 
J
Negative sign means force is opposite to the motion.
(16) (b) As the particle is moving in a circle, so
2
2
mvk
r
r
=
 Now K.E = 
1
2
 mv
2 
 = 
k
2r
36
DPP/ P 12
Now as F = – 
dU
dr
Þ P .E, U = – 
r
Fdr
¥
ò
= 
r
2
k
r
¥
æö
+
ç÷
èø
ò dr  = – 
k
r
So total energy   = U + K.E = – 
k kk
r 2r 2r
+ =-
Negative energy means that particle is in bound state .
(17) (c)  Let the mass of the person is m
Work done, W = P .E at height h above the earth surface
        = (M + m) gh
or  4900  = (M + 10) 9.8 × 10 or  M = 40 kg
(18) (b) As the rod is kept in vertical position the shift in the
ce ntre o f gra vity is e qua l to the ha lf the le ngth = l/2
Work done W = mgh = mg 
2
l
= 20 × 9.8 × 
4
2
= 392 J
(19) (a) We know that the increase in the potential energy
DU = GmM 
11
R R'
éù
-
êú
ëû
According to question R' = R + R = 2R
DU = GMm 
11
R 2R
éù
-
êú
ëû
 = 
GMm
2R
(20) (c) In first case, W
1
 = 
1
2
 m(v
1
)
2
 + mgh
=  
1
2
 m(12)
2
 + m × 10 × 12
= 72 m + 120 m = 192 m
and in second case,W
2
 = mgh = 120 m
The percentage of energy saved
= 
192m 120m
192m
-
 × 100 = 38%
(21) (c) Given that,  U (x) = 
126
ab
xx
-
We know F = – 
du
dx
 = (–12) a x
–13
 – (– 6b) x
–7
 = 0
or     
7 13
6b 12a
xx
=
or  x
6
 = 12a/6b = 2a/b  or x =
1/6
2a
b
æö
ç÷
èø
(22) (a) W = 0
(23) (b)
(1) There will be an increase in potential energy of the
system if work is done upon the system by a conservative
force.
(2) The work done by the external forces on a system equals
the change in total energy
(24) (a)
(1) The work done by all forces equal to change in kinetic
energy
(2) The work done by conservative forces equal to change
in potential energy
(3) The work done by external and nonconservative forces
equal to change in total energy
(25) (b), (26) (b), 27. (c)
For vertical block
mg = kx + 2T ....... (1)
For horizontal block
T = k (2x) ....... (2)
From eq. (1) and eq. (2)
mg
x
5k
= = 0.2m
\ Extension of vertical spring = 0.2m
Extension of horizontal spring  = 2x = 0.4m
From conservation of energy
mgx = 
2 2 22
1 1 11
kx k(2x) mv m(2v)
2 2 22
+ ++
22
33
mgx kx mv
22
=+
2
73
mgx mv
102
=
7
v gx
15
=
Required speed = 2v = 1.9 m/s
(28) (d) Statement – 1 is true but statement – 2 is false.
(29) (a) Work done by action reaction force may be zero only if
displacement of both bodies are same.
(30) (b) Both statements are true and independent.