Work, Energy and Power Practice Questions - DPP for JEE

 Page 1


1. (b) k = 5 × 10
3
 N/m
 
2. (a) Given: Mass of particle, M = 10g = 
radius of circle R = 6.4 cm
Kinetic energy E of particle = 8 × 10
–4
J
acceleration a
t
 = ?
 = E
?  = 8 × 10
–4
? v
2
 = 16 × 10
–2
? v = 4 × 10
–1
 = 0.4 m/s
Now, using
v
2
 = u
2
 + 2a
t
s (s = 4pR)
(0.4)
2
 = 0
2
 + 2a
t
 
? a
t
 = (0.4)
2
 ×  = 0.1 m/s
2
3. (b) We know that  F × v = Power
   where c = constant
 
Page 2


1. (b) k = 5 × 10
3
 N/m
 
2. (a) Given: Mass of particle, M = 10g = 
radius of circle R = 6.4 cm
Kinetic energy E of particle = 8 × 10
–4
J
acceleration a
t
 = ?
 = E
?  = 8 × 10
–4
? v
2
 = 16 × 10
–2
? v = 4 × 10
–1
 = 0.4 m/s
Now, using
v
2
 = u
2
 + 2a
t
s (s = 4pR)
(0.4)
2
 = 0
2
 + 2a
t
 
? a
t
 = (0.4)
2
 ×  = 0.1 m/s
2
3. (b) We know that  F × v = Power
   where c = constant
 
  
4. (a) When ball collides with the ground it loses its 50%
of energy
? ? 
or 
or, 
or, 4gh = 
? v
0
 = 20ms
–1
5. (c) Work done in stretching the rubber-band by a distance dx is
dW = F dx = (ax + bx
2
)dx
Integrating both sides,
6. (b) We know that ?U = – W for conservative forces
?U =  or ?U = 
U
(x)
 – U
(0)
 = 
Given U
(0)
 = 0
Page 3


1. (b) k = 5 × 10
3
 N/m
 
2. (a) Given: Mass of particle, M = 10g = 
radius of circle R = 6.4 cm
Kinetic energy E of particle = 8 × 10
–4
J
acceleration a
t
 = ?
 = E
?  = 8 × 10
–4
? v
2
 = 16 × 10
–2
? v = 4 × 10
–1
 = 0.4 m/s
Now, using
v
2
 = u
2
 + 2a
t
s (s = 4pR)
(0.4)
2
 = 0
2
 + 2a
t
 
? a
t
 = (0.4)
2
 ×  = 0.1 m/s
2
3. (b) We know that  F × v = Power
   where c = constant
 
  
4. (a) When ball collides with the ground it loses its 50%
of energy
? ? 
or 
or, 
or, 4gh = 
? v
0
 = 20ms
–1
5. (c) Work done in stretching the rubber-band by a distance dx is
dW = F dx = (ax + bx
2
)dx
Integrating both sides,
6. (b) We know that ?U = – W for conservative forces
?U =  or ?U = 
U
(x)
 – U
(0)
 = 
Given U
(0)
 = 0
U
(x)
 = 
This is the equation of a parabola, which is symmetric to U-axis in
negative direction.
7. (d) As we know power P = 
? w = Pt =  mv
2
So, v = 
Hence, acceleration 
Therefore, force on the particle at time ‘t’
= ma = 
8. (a)    (By condition of linear momentum)
...... (i)
Also  
   ...... (ii)
From (i) and (ii),   ? 
9. (b) As we know work done in stretching spring
where k = spring constant
x = extension
Case (a) If extension (x) is same,
Page 4


1. (b) k = 5 × 10
3
 N/m
 
2. (a) Given: Mass of particle, M = 10g = 
radius of circle R = 6.4 cm
Kinetic energy E of particle = 8 × 10
–4
J
acceleration a
t
 = ?
 = E
?  = 8 × 10
–4
? v
2
 = 16 × 10
–2
? v = 4 × 10
–1
 = 0.4 m/s
Now, using
v
2
 = u
2
 + 2a
t
s (s = 4pR)
(0.4)
2
 = 0
2
 + 2a
t
 
? a
t
 = (0.4)
2
 ×  = 0.1 m/s
2
3. (b) We know that  F × v = Power
   where c = constant
 
  
4. (a) When ball collides with the ground it loses its 50%
of energy
? ? 
or 
or, 
or, 4gh = 
? v
0
 = 20ms
–1
5. (c) Work done in stretching the rubber-band by a distance dx is
dW = F dx = (ax + bx
2
)dx
Integrating both sides,
6. (b) We know that ?U = – W for conservative forces
?U =  or ?U = 
U
(x)
 – U
(0)
 = 
Given U
(0)
 = 0
U
(x)
 = 
This is the equation of a parabola, which is symmetric to U-axis in
negative direction.
7. (d) As we know power P = 
? w = Pt =  mv
2
So, v = 
Hence, acceleration 
Therefore, force on the particle at time ‘t’
= ma = 
8. (a)    (By condition of linear momentum)
...... (i)
Also  
   ...... (ii)
From (i) and (ii),   ? 
9. (b) As we know work done in stretching spring
where k = spring constant
x = extension
Case (a) If extension (x) is same,
So, W
P
 > W
Q
( K
P
 > K
Q
)
Case (b) If spring force (F) is same 
So, W
Q
 > W
P
10. (c) Just before impact, energy
E = mgh = 10mg          ... (1)
Just after impact
Hence,  mgh
1
 = E
1
  
(from given figure)
mgh
1
 = 0.75 mg (10)
h
1
 = 7.5m
11. (a) Given, h = 60m, g = 10 ms
–2
,
Rate of flow of water = 15 kg/s
? Power of the falling water
= 15 kgs
–1
 × 10 ms
–2
 × 60 m = 900 watt.
Loss in energy due to friction
? Power generated by the turbine
= ( 9000 – 900) watt = 8100 watt = 8.1 kW
12. (c)
Page 5


1. (b) k = 5 × 10
3
 N/m
 
2. (a) Given: Mass of particle, M = 10g = 
radius of circle R = 6.4 cm
Kinetic energy E of particle = 8 × 10
–4
J
acceleration a
t
 = ?
 = E
?  = 8 × 10
–4
? v
2
 = 16 × 10
–2
? v = 4 × 10
–1
 = 0.4 m/s
Now, using
v
2
 = u
2
 + 2a
t
s (s = 4pR)
(0.4)
2
 = 0
2
 + 2a
t
 
? a
t
 = (0.4)
2
 ×  = 0.1 m/s
2
3. (b) We know that  F × v = Power
   where c = constant
 
  
4. (a) When ball collides with the ground it loses its 50%
of energy
? ? 
or 
or, 
or, 4gh = 
? v
0
 = 20ms
–1
5. (c) Work done in stretching the rubber-band by a distance dx is
dW = F dx = (ax + bx
2
)dx
Integrating both sides,
6. (b) We know that ?U = – W for conservative forces
?U =  or ?U = 
U
(x)
 – U
(0)
 = 
Given U
(0)
 = 0
U
(x)
 = 
This is the equation of a parabola, which is symmetric to U-axis in
negative direction.
7. (d) As we know power P = 
? w = Pt =  mv
2
So, v = 
Hence, acceleration 
Therefore, force on the particle at time ‘t’
= ma = 
8. (a)    (By condition of linear momentum)
...... (i)
Also  
   ...... (ii)
From (i) and (ii),   ? 
9. (b) As we know work done in stretching spring
where k = spring constant
x = extension
Case (a) If extension (x) is same,
So, W
P
 > W
Q
( K
P
 > K
Q
)
Case (b) If spring force (F) is same 
So, W
Q
 > W
P
10. (c) Just before impact, energy
E = mgh = 10mg          ... (1)
Just after impact
Hence,  mgh
1
 = E
1
  
(from given figure)
mgh
1
 = 0.75 mg (10)
h
1
 = 7.5m
11. (a) Given, h = 60m, g = 10 ms
–2
,
Rate of flow of water = 15 kg/s
? Power of the falling water
= 15 kgs
–1
 × 10 ms
–2
 × 60 m = 900 watt.
Loss in energy due to friction
? Power generated by the turbine
= ( 9000 – 900) watt = 8100 watt = 8.1 kW
12. (c)
 Total time taken by the body in coming to rest
=  
13. (d) When C strikes A
 ( = velocity of A)
............. (i)
 = 
(When A  and  B Block attains K.E.)
......... (ii)