5 ข้อสอบ Physics Student Pilot Work & Energy
Question 1 — Work
A ground vehicle pulls an aircraft with a constant horizontal force of 2,400 N over a distance of 50 m. The force acts in the same direction as the displacement.
How much work is done by the pulling force?
A. 48,000 J
B. 60,000 J
C. 100,000 J
D. 120,000 J
E. 240,000 J
Answer: D. 120,000 J
Explanation:
Work = Force × Distance
W = F × s
W = 2,400 × 50
W = 120,000 J
Shortcut:
ถ้าแรงกับการเคลื่อนที่ไปในทิศเดียวกัน ใช้ได้ทันทีว่า
W = F × s
Question Type: Work
Difficulty: Easy
Question 2 — Kinetic Energy
An aircraft of mass 1,500 kg increases its speed from 20 m/s to 40 m/s.
What is the increase in kinetic energy?
A. 600 kJ
B. 750 kJ
C. 900 kJ
D. 1,000 kJ
E. 1,200 kJ
Answer: C. 900 kJ
Explanation:
ΔKE = ½m(v² − u²)
ΔKE = ½(1,500)(40² − 20²)
= 750(1,600 − 400)
= 750 × 1,200
= 900,000 J
= 900 kJ
Shortcut:
อย่าใช้ ½m(v − u)²
สูตรที่ถูกต้องคือ
ΔKE = ½m(v² − u²)
Question Type: Kinetic Energy
Difficulty: Medium
Question 3 — Gravitational Potential Energy
A 12 kg piece of aircraft equipment is lifted vertically through a height of 8 m.
Assume g = 10 m/s².
How much gravitational potential energy does it gain?
A. 96 J
B. 120 J
C. 480 J
D. 800 J
E. 960 J
Answer: E. 960 J
Explanation:
PE = mgh
PE = 12 × 10 × 8
PE = 960 J
Shortcut:
Potential Energy ใช้ความสูงในแนวดิ่ง
PE = mgh
Question Type: Gravitational Potential Energy
Difficulty: Easy–Medium
Question 4 — Work-Energy Theorem
An aircraft of mass 2,000 kg starts from rest. The engine provides a constant thrust of 5,000 N while the total resistive force is 1,000 N.
If the aircraft travels 100 m, what is its speed at the end of the distance?
A. 10 m/s
B. 15 m/s
C. 20 m/s
D. 25 m/s
E. 30 m/s
Answer: C. 20 m/s
Explanation:
Step 1: Find net force.
Fnet = 5,000 − 1,000
Fnet = 4,000 N
Step 2: Find net work.
Wnet = Fnet × s
Wnet = 4,000 × 100
Wnet = 400,000 J
Step 3: Use the Work-Energy Theorem.
Wnet = ΔKE
Since the aircraft starts from rest:
400,000 = ½(2,000)v²
400,000 = 1,000v²
v² = 400
v = 20 m/s
Shortcut:
For an object starting from rest:
Fnet × s = ½mv²
Question Type: Work-Energy Theorem
Difficulty: Medium–Hard
Question 5 — Conservation of Mechanical Energy
A 3 kg object is released from rest at a height of 20 m above the ground. It falls without air resistance.
Assume g = 10 m/s².
What is its speed when it reaches a height of 5 m above the ground?
A. 10.0 m/s
B. 12.5 m/s
C. 15.0 m/s
D. 17.3 m/s
E. 20.0 m/s
Answer: D. 17.3 m/s
Explanation:
The object falls through:
Δh = 20 − 5
Δh = 15 m
Using conservation of mechanical energy:
mgh = ½mv²
Mass cancels:
gh = ½v²
10 × 15 = ½v²
150 = ½v²
v² = 300
v = √300
v ≈ 17.3 m/s
Shortcut:
If an object starts from rest and there is no friction or air resistance:
v = √(2gΔh)
Notice that mass does not affect the final speed.
Question Type: Conservation of Mechanical Energy
Difficulty: Hard
สูตรสำคัญที่ควรจำ
W = F × s × cosθ
KE = ½mv²
PE = mgh
Wnet = ΔKE
Mechanical Energy:
KE₁ + PE₁ = KE₂ + PE₂
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