## Energy, Work and Power

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## Energy, Work, Power (O Level)

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Question 1 |

A block of ice is sliding at a constant speed before moving up a smooth slope as shown. What are the changes in its kinetic exergy and gravitational potential energy as it is sliding up the slope?

Kinetic Energy Gravitational Potential
energy
Increase increase
| |

Increase decrease | |

decrease increase | |

decrease decrease |

Question 1 Explanation:

As the ice is sliding upwards, there is a gain of height. Kinetic energy of the ice is used to convert to the gravitational potential energy of the ice. KE therefore decreases while PE increases.

Question 2 |

A car is accelerating up a slope. What are the changes in its kinetic energy and gravitational potential energy as it is moving up the slope?

Kinetic Energy Gravitational Potential
energy
Increase increase
| |

Increase decrease | |

decrease increase | |

decrease decrease |

Question 2 Explanation:

The keywords in this question are “accelerating” and “up a slope”. “Accelerating” indicates that the car is increasing speed and thus the KE must increase. “Up a slope” indicates that the car is increasing height and thus the PE must increase. The energy to increase KE and PE come from the petrol which is not mentioned in the question.

Question 3 |

A box is being pushed horizontally on a smooth surface by a 10 N force for 6 m. What is the work done on the box by the 10 N force?

0.6 J | |

1.7 J | |

16 J | |

60 J |

Question 3 Explanation:

Work done = force x distance moved in the direction of the force
= 10 x 6 = 60 J

Question 4 |

A box of mass 2 kg is being pushed horizontally on a smooth surface by a 10 N force for 6 m. What is the work done on the box by the 10 N force?

12 J | |

48 J | |

60 J | |

120 J |

Question 4 Explanation:

Work done = force x distance moved in the direction of the force
= 10 x 6 = 60 J
Note that the mass is ignored as the question is only interested in work done by the 10 N force and not the weight.

Question 5 |

A box of mass 2 kg is being pushed horizontally on a rough surface by a 10 N force for 6 m. Given that the friction between the box and the surface is 4 N, what is the work done on the box by the 10 N force?

24 J | |

36 J | |

60 J | |

120 J |

Question 5 Explanation:

Work done = force x distance moved in the direction of the force
= 10 x 6 = 60 J
Note that the question is only asking for the work done by the 10 N force. All other work done is not considered.

Question 6 |

A box of mass 2 kg is being pushed horizontally on a smooth surface by a 10 N force for 6 m. What is the work done on the box by the weight of the box? (g = 10 N/kg)

0 J | |

12 J | |

60 J | |

120 J |

Question 6 Explanation:

Weight of the box = mg = 2 x 10 = 20 N
Work done = force x distance moved in the direction of the force

Question 7 |

A box of mass 2 kg is being pushed horizontally by a 10 N force for 60 cm along the floor. During the first 20 cm, the friction between the box and the floor is 4 N, and the friction for the rest of the distance is negligible. What is the work done on the box by the 10 N force?

2 J | |

3 J | |

4 J | |

6 J |

Question 7 Explanation:

Work done = force x distance moved in the direction of the force
= 10 x 0.6 = 6 J

Question 8 |

A box of mass 2 kg is moving at a constant speed of 3 m/s. What is the kinetic energy of the box?

2 J | |

3 J | |

9 J | |

18 J |

Question 8 Explanation:

Kinetic Energy = ½ m x v x v
= ½ x 2 x 3 x 3 = 9 J

Question 9 |

A box of mass 500 g is moving at a constant speed of 72 km/h. What is the kinetic energy of the box?

5 J | |

100 J | |

1296 J | |

1296000 J |

Question 9 Explanation:

72 km/h = (72 x 1000 m) / (3600 s)
= 20 m/s
Kinetic Energy = ½ m x v x v
= ½ x 0.5 x 20 x 20 = 100 J

Question 10 |

A box of mass 2 kg increases its speed from 3 m/s to 7 m/s. What is the increase in kinetic energy?

4 J | |

16 J | |

40 J | |

32 J |

Question 10 Explanation:

Initial Kinetic Energy = ½ m x v x v
= ½ x 2 x 3 x 3 = 9 J
Final Kinetic Energy = ½ m x v x v
= ½ x 2 x 7 x 3 = 49 J
Increase in Kinetic Energy = 49 J – 9 J
= 40 J

Question 11 |

A box of mass 2 kg is lifted vertically upwards by 3 m. Given that the acceleration due to gravity is 10 m/s

^{2}, what is the gravitational potential energy gained by the box?0.6 J | |

6.7 J | |

15 J | |

60 J |

Question 11 Explanation:

Gravitational potential energy = mgh
= 2 x 10 x 3
= 60 J

Question 12 |

A box of mass 2 kg is lifted diagonally from point A to point B as shown. Given that the acceleration due to gravity is 10 m/s

^{2}, what is the gravitational potential energy gained by the box?20 J | |

60 J | |

80 J | |

100 J |

Question 12 Explanation:

Gravitational potential energy = mgh
= 2 x 10 x 3
= 60 J
Note that ‘mgh’ is actually a work done by lifting the weight of the object. The weight of the object is ‘mg’ and the distance moved by the weight in the direction of the lifting force is h.

Question 13 |

There are 3 paths leading to the top of the hill as shown. Assuming that the friction of the ground is negligible, which of the following statements is true?

Path C requires most energy to reach the top. | |

Path B requires the least energy to reach the top. | |

Path B requires the least energy to reach the top. | |

All three paths require same amount of energy to reach the top. |

Question 13 Explanation:

The work done to reach the top of the hill is the same for all three routes when there is no friction along the ways. The height achieved by the three routes is the same. If there is friction, taking the longer route will encounter a longer stretch of rough surface and will require more energy to overcome the friction.

Question 14 |

A box of mass 2 kg moves up a ramp from point A to point B at a constant speed. Given that the surfaces are smooth (frictionless), what is the gain in the potential energy by the box? (g = 10 N/kg)

80 J | |

100 J | |

200 J | |

220 J |

Question 14 Explanation:

Gravitational potential energy = mgh
= 2 x 10 x 4
= 80 J
When an object is moving horizontally, its height does not increase and there is no increase in potential energy.

Question 15 |

A pendulum bob pivots at a point P, starts swinging from point S. Assuming that there is no loss os energy to the surrounding, at which position will the bob next come to a momentary rest?

A | |

B | |

C | |

D |

Question 15 Explanation:

Principle of conservation of energy states that energy cannot be destroyed or created. It can only change from one form to another.
When there is no energy lost to the surrounding, the bob will conserve its energy and go to position C. At position C, the bob’s total energy is the same as the total energy at the initial state.

Question 16 |

Three different stones are being lifted from the ground level to different heights on different planets as given below.
Case 1: A 250 g stone lifted to 10 m on Earth.
Case 2: A 6 kg stone lifted to 2 m on the moon
Case 3: A 1 kg stone lifted to 1 m on Jupiter
Given that the acceleration due to gravity on Earth, the Moon and Jupiter are 10 m/s

^{2}, 2 m/s^{2}, 25 m/s^{2}respectively, which of the following statements is true?More energy is required for case 2 than case 1. | |

More energy is required for case 1 than case 3. | |

Case 1 and case 2 require the same amount of energy | |

Case 1 and case 3 require the same amount of energy |

Question 16 Explanation:

Work done required for case A
= mgh
= 0.25 x 10 x 10
= 25 J
Work done required for case B
= mgh
= 6 x 2 x 2
= 24 J
Work done required for case C
= mgh
= 1 x 25 x 1
= 25 J

Question 17 |

A box is pushed horizontally on a smooth surface by a 10 N force for 6 m. What is the increase in the kinetic energy of the box?

0.6 J | |

1.7 J | |

16 J | |

60 J |

Question 17 Explanation:

The energy used to increase the kinetic energy of the box is given by the work done on it. The box increases in speed due to the push. The longer the force applies on the box, the higher will be the speed gain by the box.
Increase in kinetic energy = work done on the box
= force x distance
= 10 x 6
= 60 J

Question 18 |

A 1.2 kg box is being pushed horizontally on a smooth surface by a 10 N force for 6 m. Given that the box is initially at rest, what is the speed of the box at the end of the 6 m?

2.0 m/s | |

10 m/s | |

50 m/s | |

100 m/s |

Question 18 Explanation:

Due to conservation of energy,
increase in kinetic energy = work done on box
½ m x v x v = force x distance
½ (1.2) x v x v = 10 x 6
v = 10 m/s

Question 19 |

A 6 kg box is being pushed horizontally on a smooth surface by a 10 N force for 6 m. Given that the box has an initial speed of 4m/s, what is the speed of the box at the end of the 6 m?

2.0 m/s | |

4.5 m/s | |

6.0 m/s | |

10 m/s |

Question 19 Explanation:

Due to conservation of energy,
increase in kinetic energy = work done on box
(½ x m x v x v) - (½ m x u x u) = force x distance
(½ (6) x v x v) - (½ (6) x 4 x 4) = 10 x 6
v = 6 m/s

Question 20 |

A 6 kg box is being pushed horizontally on a surface by a 10 N force for 6 m. Given that the box has an initial speed of 4m/s, what is the speed of the box at the end of the 6 m?

2.0 m/s | |

4.5 m/s | |

5.7 m/s | |

10 m/s |

Question 20 Explanation:

Due to conservation of energy,
input energy = output energy
Work done on box = increase in kinetic energy +
work done against friction
force x distance = (½ x m x v x v) - (½ m x u x u) + frictional force x distance travelled
10 x 6 = (½ (6) x v x v) - (½ (6) x 4 x 4) + 2 x 6
v = 5.7 m/s

Question 21 |

A 6 kg box is being pushed horizontally on a surface by a 10 N force for 6 m. Given that the box has an initial speed of 4 m/s and a final speed of 5 m/s at the end of the 6 m, what is the friction between the box and the surface?

2.0 N | |

3.5 N | |

5.5 N | |

6.0 N |

Question 21 Explanation:

C. Due to conservation of energy,
input energy = output energy
Work done on box = increase in kinetic energy +
work done against friction
force x distance = (½ x m x v x v) - (½ m x u x u) + frictional force x distance travelled
10 x 6 = (½ (6) x 5 x 5) - (½ (6) x 4 x 4) + F x 6
F = 5.5 N

Question 22 |

A 6 kg box is being pulled up a ramp by a force F at a constant speed as shown. Given that the friction along the surface is 4 N, what is the force F required to bring the box to the top of the ramp? (g = 10 N/kg)

4 N | |

40 N | |

44 N | |

64 N |

Question 22 Explanation:

Due to conservation of energy,
input energy = output energy
Work done on box = increase in potential energy +
work done against friction
force x distance = mgh - frictional force x distance travelled
F x 6 = (6)(10)(4) - 4 x 6
F = 44 N

Question 23 |

A box of 6 kg mass is being pulled up a ramp by a force F from rest as shown. Given that the friction along the surface is 4 N and the final speed of the box on top of the ramp is 2 m/s, what is the force F required to bring the box to the top of the ramp? (g = 10 N/kg)

4 N | |

42 N | |

46 N | |

64 N |

Question 23 Explanation:

Due to conservation of energy,
input energy = output energy
Work done on box = increase in kinetic energy +
work done against friction +
increase in potential energy
force x distance = (½ x m x v x v) - (½ m x u x u) + frictional force x distance travelled + mgh
F x 6 = (½ (6) x 2 x 2) - (½ (60) x 04 x 04) + (4 x 6) + (6 x 10 x 4)
F = 46 N

Question 24 |

A box of 6 kg mass is being pulled by a 50-N force from an initial speed of 4 m/s at the foot of the ramp to a final speed of 6 m/s at the top of the ramp. What is the magnitude of friction along the ramp? (g = 10 N/kg)

0 N | |

4 N | |

12 N | |

50 N |

Question 24 Explanation:

Due to conservation of energy,
input energy = output energy
Work done on box = increase in kinetic energy + work done against friction + increase in potential energy
force x distance = (½ x m x v x v) - (½ m x u x u) + (frictional force x distance travelled) + mgh
50 x 6 = (6) x 6 x 6) - (6 x 4 x 4) + (F x 6) + (6 x 10 x 4)
F = 0 N

Question 25 |

A box of 6 kg mass is being pulled by a 8-N force as shown, from an initial speed of 4 m/s at the foot of the ramp to a final speed of 6 m/s at the top of the ramp. Given that the friction along the ramp is 2 N, what is the mass of the box? (g = 10 N/kg)

0.2 kg | |

0.6 kg | |

1.2 kg | |

2.4 kg |

Question 25 Explanation:

Due to conservation of energy,
input energy = output energy
Work done on box = increase in kinetic energy + work done against friction + increase in potential energy
force x distance = (½ x m x v x v) - (½ m x u x u) + (frictional force x distance travelled) + mgh
8 x (5 ÷ cos 60) = (1/2)(M)(6 x 6) - (1/2)(M)(4 x 4) + (2) (5 ÷ cos 60) + (M)(10)(5 x tan 60)
M = 0.62 kg

Question 26 |

A box of 2 kg mass has an initial speed of 10 m/s at the foot of the ramp. Given that the friction along the ramp is 2 N, calculate the maximum height h that it can reach? (g = 10 N/kg)

2.8 m | |

4.5 m | |

5.0 m | |

43.3 m |

Question 26 Explanation:

Due to conservation of energy,
input energy = output energy
Decrease in kinetic energy = work done against friction + increase in potential energy
(½ x m x u x u) - (½ m x v x v) = (frictional force x distance travelled) + mgh
(1/2)(2)(10 x 10) - (1/2)(2)(0 x 0) = (2) (h ÷ sin 60) + (2)(10)(h)
h = 4.48 m

Question 27 |

A box of mass 2 kg has an initial speed of 10 m/s at the foot of the ramp. Given that the friction along the ramp is 2 N, calculate the height

*h*that it reaches when the speed if the box is 5 m/s. (g = 10 N/kg)2.5 m | |

3.4 m | |

4.5 m | |

6.0 m |

Question 27 Explanation:

Due to conservation of energy,
input energy = output energy
Decrease in kinetic energy = work done against friction + increase in potential energy
(½ x m x u x u) - (½ m x v x v) = (frictional force x distance travelled) + mgh
(1/2)(2)(10 x 10) - (1/2)(2)(5 x 5) = (2) (h ÷ sin 60) + (2)(10)(h)
h = 3.36 m

Question 28 |

A box of mass 2 kg slides down from the top of the ramp with an initial speed of 0 m/s. Given that the friction along the ramp is 3.2 N, calculate the speed of the box when it reaches the foot of the ramp. (g = 10 N/kg)

2.8 m/s | |

4.5 m/s | |

8.0 m/s | |

8.9 m/s |

Question 28 Explanation:

Due to conservation of energy,
input energy = output energy
Decrease in potential energy = work done against friction + increase in kinetic energy
[Length of the slope]^2 = (3 x 3) + (4 x 4)
Length of the slope = 5 m
mgh = (½ x m x u x u) - (½ m x v x v) + (frictional force x distance travelled)
(2)(10)(4) = (1/2)(2)(v x v) - (1/2)(2)(0 x 0) + (3.2) (5)
v = 8 m/s

Question 29 |

A box is being pushed horizontally on a smooth surface by a 10 N force for 6 m in 3 s. What is the power acting on the box during this period?

5 W | |

18 W | |

20 W | |

600 W |

Question 29 Explanation:

Work done = force x distance travelled in the direction of the force
= 10 x 6 = 60 J
Power = Work done ÷ time
= 60 ÷ 3 = 20 W

Question 30 |

A man lifts a box of mass 2 kg vertically upwards by 3 m. Given that he takes 6 s to do the work, what is the power generated by the man? (g = 10 m/s

^{2})1 W | |

6 W | |

10 W | |

360 W |

Question 30 Explanation:

Work done = force x distance travelled in the direction of the force
= 2 x 10 x 3 = 60 J
Power = Work done ÷ time
= 60 ÷ 6 = 10 W

Question 31 |

A bullet hits a target board at a speed of 300 m/s. Given that the bullet is finally embedded 3.0 cm into the target board and the mass of the bullet is 15 g, what is the average resistive force acting on the bullet while it is penetrating into the target board?

225 N | |

2250 N | |

22500 N | |

225000 N |

Question 31 Explanation:

15 g = 0.015 kg, 3 cm = 0.03 m
Due to conservation of energy,
input energy = output energy
Decrease in kinetic energy = work done against friction
(½ x m x u x u) - (½ m x v x v) = (frictional force x distance travelled)
(1/2)(0.015)(300 x 300) - (1/2)(0.015)(0 x 0) = F x (0.03)
F = 22500 N

Question 32 |

A 100 g mass drops from a 10 m building. What is the kinetic energy of the mass when it is 2 m away from the floor? (g = 10 m/s

^{2})0 J | |

2 J | |

8 J | |

10 J |

Question 32 Explanation:

Total energy when the mass is at 10 m = mgh = (0.1)(10)(10) = 10 J
Total energy of mass at 2 m
= gravitation potential energy + kinetic energy
= (0.1)(10)(2) + kinetic energy
Due to conservation of energy, total energy when the mass is at 2 m = 10 J
10 = (0.1)(10)(2) + kinetic energy
v = 8 J

Question 33 |

A 100 g mass drops from a 10 m building. What is the speed of the mass when it is 4 m away from the floor? (g = 10 m/s

^{2})4 m/s | |

6 m/s | |

8 m/s | |

11 m/s |

Question 33 Explanation:

Total energy when the mass is at 10 m
= mgh
= (0.1)(10)(10) = 10 J
Total energy of mass at 4 m
= gravitation potential energy + kinetic energy
= (0.1)(10)(4) + (1/2)(0.1)(v x v)
Due to conservation of energy, total energy when the mass is at 4 m = 10 J
10 = (0.1)(10)(4) + (1/2)(0.1)(v x v)
v = 11.0 m/s

Question 34 |

A 1000 kg motorcar travels at a constant speed of 30 m/s along a road with frictional force of 100 N. What is the power generated by the motorcar? (g = 10 m/s

^{2})3000 W | |

4500 W | |

300000 W | |

450000 W |

Question 34 Explanation:

Power
= Work done ÷ time taken
= Force x distance ÷ time taken
= Force x (distance ÷ time taken)
= Force x constant speed
= 100 x 30 = 3000 W

Question 35 |

A motor car of mass 500 kg generates a power of 10000 W. How much time does the motorcar need to accelerate from a speed of 10 m/s to 20 m/s?

2.5 s | |

5.0 s | |

7.5 s | |

10 s |

Question 35 Explanation:

Due to conservation of energy,
input energy = output energy
Power x time = increase in kinetic energy
P x t = (1/2)[(m x v x v) - (m x u x u)]
10000t = (1/2)(500)(20 x 20) – (500)(10 x 10)
t = 7.5 s

Question 36 |

A motor car of mass 500 kg generates a power of 10000 W. Given that the total resistance on the motorcar need to accelerate from a speed of 10 m/s to 20 m/s?

6.3 s | |

8.3 s | |

9.2 s | |

10.7 s |

Question 36 Explanation:

Distance travelled = area under graph
= (1/2) x (10 + 20) x t = 15t
Due to conservation of energy,
input energy = output energy
Power x time = work done against friction + increase in kinetic energy
P x t = force x distance + (1/2)[(m x v x v) – (m x u x u)]
10000t = (200)(15t) + (1/2)[(500)(20 x 20) – (500)(10 x 10)]
t = 10.7 s

Question 37 |

A 100 w spot light has an efficiency of 40%. How much light energy can the spotlight produce in an hour?

144 kJ | |

216 kJ | |

360 kJ | |

720 kJ |

Question 37 Explanation:

1 hour = 3600 s
Energy = Power x time
= 100 x 3600
= 360000 J
Efficiency of bulb = 40% (means 40% of energy is useful and the rest of 60% converts to heat)
Light energy = 0.4 x 360000
= 144 kJ

Question 38 |

A worm of mass 2 g crawls up a flight of steps from point A to point B as shown. What is the work done by the worm? (g = 10 N/Kg)

1.2 mJ | |

3.6 mJ | |

0.12 J | |

120 J |

Question 38 Explanation:

2 g = 0.002 kg, 2 cm = 0.02 m
Input energy by worm = increase in gravitational potential energy
Input energy by worm = (0.002)(10)(0.02 + 0.02 + 0.02)
= 0.0012 J
= 1.2 mJ

Question 39 |

A worm of mass 2 g crawls up a flight of steps from point A to point B as shown. Given that the friction of the floor on the worm is 0.001 N, what is the work done by the worm? (g = 10 N/Kg).

0.12 mJ | |

0.18 mJ | |

1.20 mJ | |

1.38 mJ |

Question 39 Explanation:

2 g = 0.002 kg, 2 cm = 0.02 m, 3 cm = 0.03 m
Input energy by worm = increase in gravitational potential energy + work done against friction
Input energy by worm = (0.002)(10)(0.02 + 0.02 + 0.02) + (0.001)(0.03 + 0.02 + 0.03 + 0.02 + 0.03 + 0.02 + 0.03)
= 0.00138 J
= 1.38 mJ

Question 40 |

The speed time graph below shows the motion of a motorcar on a level road within a 10 s duration. Given that the mass of the motorcar is 500 kg and the frictional force along the road is 120 N, what is the power of the motorcar?

3200 W | |

4160 W | |

32000 W | |

41600 W |

Question 40 Explanation:

Distance travelled within 10 s
= area under graph
= (1/2)(4 + 12)(10) = 80 m
Input energy = Output energy
P x t = (½)(m x v x v) – (1/2)(m x u x u) + frictional force x distance travelled
P x (10) = (1/2)(500)(12 x 12) – (1/2)(500)(4 x 4) + (120)(80)
P = 4160 W

Question 41 |

A mass of 2 kg is suspended by a 1.0 m string and is set free at point A as shown. Neglecting air resistance, what is the speed of the mass as it passes through point B? (g = 10 m/s)

1.5 m/s | |

2.3 m/s | |

3.2 m/s | |

4.1 m/s |

Question 41 Explanation:

Total energy initially at the top = mgh
= 2 x 10 x 0.5 = 10 J
Total energy finally at the bottom
= (1/2)(m x v x v)
= (1/2)(2 x v x v) = (v x v)
By the principle of conservation of energy
10 = (v x v)
v = 3.16 m/s

Question 42 |

A mass of mass 3 kg is being pulled along horizontal bench at a constant speed of 5 m by a 10-N horizontal force. What is the work done against friction in 4 s?

0 J | |

200 J | |

600 J | |

1000 J |

Question 42 Explanation:

Frictional force = 10 N
Distance travelled in 4 seconds
= v x t = 5 x 4 = 20 m/s
Work done against friction
= F x d = 10 x 20 = 200 J

Question 43 |

A block of mass

*m*slides from the rest down smooth incline of length*L*. As a result, it moves through a vertical height*y*and a horizontal distance*x*. When it reaches the bottom of the incline, what is its kinetic energy? (Take acceleration due to gravity as g)mgL | |

mgy | |

(0.5)(m)(x^2) | |

(0.5)(m)(y^2) |

Question 43 Explanation:

Base on conservation of energy principle
KE = PE
KE = mgy

Question 44 |

A car of mass

*m*has an engine which can deliver power*P*. What is the minimum time in which the car can be accelerated from rest to a speed*v*?mv / P | |

P / mv | |

(m x v x v) / 2P | |

2P / (m x v x v) |

Question 44 Explanation:

E = Pt
(1/2)(m x v x v) - (1/2)(m x u x u) = Pt
Since the initial speed u is 0 m/s
t = (1/2)(m x v x v) ÷ P
t = (m x v x v) ÷ 2P

Question 45 |

A block of mass 1 kg has a speed of 20 m/s at point A. It is moving towards point C. Given that the frictional force along AB is 10 N and the frictional force along BC is 5 N, what is the speed of the block when it reaches point C and what is the time taken for the process? (Take acceleration due to gravity to be 10 m/s)

Speed Time taken
15.2 m/s 0.45 s
| |

15.2 m/s 0.55 s | |

17.0 m/s 0.65 s | |

17.0 m/s 0.75 s |

Question 45 Explanation:

Total KE of the 1 kg mass
= (1/2)(m x v x v) = (1/2) x 1 x (20 x 20) = 200 J
Energy lost as heat along AB
= Frictional Force x Distance
= 10 x 3 = 30 J
Energy lost as heat along BC
= Frictional Force x Distance + mgh
= (5 x 5) + (1 x 10 x 3) = 55 J
KE remaining at point C
= 200 – 30 – 55 = 115 J
(1/2)(m x v x v) = 115
v = sqrt(115 x 2)
v = 15.2 m/s

Question 46 |

A bullet leaves the vertical barrel of a rifle and reaches a height of 720 m. What is the speed of the bullet when it leaves the barrel of the rifle to achieve this height? (Given that the acceleration due to gravity is 10 m/s)

120 m/s | |

170 m/s | |

240 m/s | |

14400 m/s |

Question 46 Explanation:

Initial energy at the bottom
= (1/2)(m x v x v) + 0 = (1/2)(m x v x v)
Final energy at the highest point
= KE + PE = 0 + mgh
= m(10)(720)
= 7200 m
Conservation of energy,
(1/2)(m x v x v) = 7200m
v x v = 14400
v = 120 m/s

Question 47 |

A cable car of mass 200 kg travels up a hill from an altitude of 1030 m to an altitude of 1150 m. During this period, the cable car moves through a total of 250 m cable length in 5 min. What is the rate of increase of potential energy?

800 W | |

1000 W | |

39500 W | |

48000 W |

Question 47 Explanation:

Energy to increase potential energy
= mgh
= 200 x 10 x (1150 – 1030)
= 240000 J
Power to increase potential energy
= 240000 ÷ (5 x 60)
= 800 W

Question 48 |

A block made from steel falls from position 1 to position 2 as shown. How much potential energy is lost during the process? (g = 10 m/s, density of steel = 8000 kg/m

^{3})12.8 MJ | |

16.0 MJ | |

25.6 MJ | |

32.0 MJ |

Question 49 |

A block resting on a smooth horizontal floor is being pushed by a constant horizontal force. Find the following ratio:

Kinetic energy gained in 1 |
: |
Kinetic energy gained in 2 |
: |
Kinetic energy gained in 3 |

1 : 2 : 3 | |

1 : 4 : 9 | |

1 : 3 : 5 | |

1 : 1 : 1 |

Question 49 Explanation:

Kinetic energy at 1st second = (1/2)(m x v x v)
Kinetic energy at 2nd second
= (1/2)(m x 2v x 2v)
= 4 x [(1/2)(m x v x v)]
Kinetic energy at 3rd second
= (1/2)(m x 3v x 3v)
= 9 x [(1/2)(m x v x v)]
Kinetic energy gained within 1st second
= (1/2)(m x v x v) – 0 = (1/2)(m x v x v)
Kinetic energy gained within 2nd second
= 4 x [(1/2)(m x v x v)] – (1/2)(m x v x v)
= 3 x [(1/2)(m x v x v)]
Kinetic energy gained within 3rd second
= 9 x [(1/2)(m x v x v)] – 4 x [(1/2)(m x v x v)]
= 5 x [(1/2)(m x v x v)]
Ratio is 1 : 3 : 5

Question 50 |

A 1 kg mass and a 5 kg mass are held at a position along the ramp as shown. Given that the friction along the ramp is 10 N, what is the speed of the 1 kg mass after the masses are released and the 1 kg mass has moved for 10 cm?

0.5 m/s | |

1.1 m/s | |

1.7 m/s | |

2.5 m/s |

Question 50 Explanation:

Input energy = output energy
Decrease in potential energy = work done against friction + increase in kinetic energy
Height decrease by 5 kg mass = 10 cm = 0.1 m
Height increase by 1 kg mass
= (10 sin 30) cm = 5 cm = 0.05 m
= (1/2)(1 + 5)(v x v) + 10 x 0.1
v = 1.08 m/s

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