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CBSE Class 9  >  Class 9 Test  >  Science New NCERT 2026-27 (New Syllabus)  >  Test: Work,Energy, and Simple Machines - 1 - Class 9 MCQ

Work,Energy, and Simple Machines - 1 - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Work,Energy, and Simple Machines - 1 (15 Questions)

You can prepare effectively for Class 9 Science Class 9 New NCERT 2026-27 (New Syllabus) with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Work,Energy, and Simple Machines - 1". These 15 questions have been designed by the experts with the latest curriculum of Class 9 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 15 minutes
  • - Number of Questions: 15

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Test: Work,Energy, and Simple Machines - 1 - Question 1
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What is the SI unit of work and energy?

Detailed Solution: Question 1

The SI unit of work and energy is the joule (J), named after James Prescott Joule. 1 J = 1 N × 1 m = 1 kg m² s⁻². It is defined as work done when a 1 N force displaces an object by 1 m in the direction of force.

Test: Work,Energy, and Simple Machines - 1 - Question 2
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A porter carries 15 kg luggage horizontally for 5 m. Work done on luggage by porter? (g = 10 m/s²)

Detailed Solution: Question 2

The porter applies upward force (= mg = 150 N) to support the luggage, but displacement is horizontal. Force ⊥ displacement → W = F × s × cos90° = 0. Zero work is done in the scientific sense.

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Test: Work,Energy, and Simple Machines - 1 - Question 3
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Which is the correct formula for kinetic energy?

Detailed Solution: Question 3

Derived using Work-Energy Theorem: W = F × s = ma × v²/2a = ½mv². So K = ½mv². Option C (without ½) is a common error students make — always include the ½.

Test: Work,Energy, and Simple Machines - 1 - Question 4
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A 0.5 kg ball is thrown upward at 10 m/s. What is its kinetic energy at the moment of throwing?

Detailed Solution: Question 4

K = ½mv² = ½ × 0.5 × (10)² = ½ × 0.5 × 100 = 25 J. At maximum height: v = 0, so KE = 0 and all energy is PE. Using v² = u² − 2gh → h = 5 m → PE = 0.5 × 10 × 5 = 25 J ✓ (energy conserved).

Test: Work,Energy, and Simple Machines - 1 - Question 5
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Work done by gravity on a 5 kg object falling 4 m is: (g = 10 m/s²)

Detailed Solution: Question 5

Gravity acts downward; displacement is also downward — same direction. W = mg × h = 5 × 10 × 4 = +200 J. Positive work because force and displacement are in the same direction. This work increases the kinetic energy of the falling object.

Test: Work,Energy, and Simple Machines - 1 - Question 6
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Which formula correctly expresses power?

Detailed Solution: Question 6

Power = Work / Time. P = W/t. SI unit is watt (W). 1 W = 1 J/s. If 100 J of work is done in 5 s, P = 20 W. Option A is wrong (that would give J², not power). Option D gives force, not power.

Test: Work,Energy, and Simple Machines - 1 - Question 7
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When does a girl do negative work on a dumbbell?

Detailed Solution: Question 7

While lowering: she applies upward force (to control descent) but displacement is downward. Force and displacement are opposite → negative work. While lifting: both force and displacement are upward → positive work. Negative work means she is removing energy from the dumbbell's motion.

Test: Work,Energy, and Simple Machines - 1 - Question 8
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Gravitational potential energy of mass m at height h is:

Detailed Solution: Question 8

U = mgh. To raise object slowly, apply F = mg. Work done = mg × h = mgh. By Work-Energy Theorem, this appears as stored PE = mgh. Valid near Earth's surface where g ≈ 10 m/s². Option D (½mgh) is incorrect — there is no ½ factor in gravitational PE.

Test: Work,Energy, and Simple Machines - 1 - Question 9
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A 10 kg object is raised to 3 m height. Its gravitational PE is: (g = 10 m/s²)

Detailed Solution: Question 9

U = mgh = 10 × 10 × 3 = 300 J. If the object is dropped from 3 m: ½mv² = 300 → v² = 60 → v ≈ 7.7 m/s at ground. This confirms energy conversion from PE to KE during free fall.

Test: Work,Energy, and Simple Machines - 1 - Question 10
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In which situation is work done by applied force equal to ZERO?

Detailed Solution: Question 10

For a satellite in circular orbit, gravitational force acts toward Earth's centre — always perpendicular to velocity and displacement. W = F × s × cos90° = 0. For all other options, the force has a component along displacement, so work ≠ 0.

Test: Work,Energy, and Simple Machines - 1 - Question 11
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If velocity is tripled, kinetic energy becomes how many times?

Detailed Solution: Question 11

K = ½mv². New KE = ½m(3v)² = 9 × ½mv² = 9 times. KE ∝ v². Doubling v → 4× KE. Tripling v → 9× KE. This explains why road accidents at high speed are far more destructive.

Test: Work,Energy, and Simple Machines - 1 - Question 12
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Mechanical Advantage is defined as:

Detailed Solution: Question 12

MA = Load / Effort. If MA > 1, a small effort overcomes a large load. MA = 3 means 100 N effort lifts 300 N load. Simple machines don't reduce total work — effort decreases but distance increases proportionally.

Test: Work,Energy, and Simple Machines - 1 - Question 13
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An inclined plane has length 5 m and height 1 m. Its mechanical advantage is:

Detailed Solution: Question 13

MA = L/h = 5/1 = 5. A load can be pushed up the ramp with only 1/5th of its weight as effort. However, effort acts over 5 m while load only rises 1 m. Work done = (load/5) × 5 m = load × 1 m. Total work is the same — only effort is reduced.

Test: Work,Energy, and Simple Machines - 1 - Question 14
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At the highest point of a vertically thrown ball, its kinetic energy is:

Detailed Solution: Question 14

At the highest point, velocity = 0 momentarily. K = ½mv² = ½m(0)² = 0. All KE has been converted to PE = mgh_max. Total ME = KE + PE = constant throughout the journey (conservation of mechanical energy, ignoring air resistance).

Test: Work,Energy, and Simple Machines - 1 - Question 15
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At the lowest point of a simple pendulum, which is correct?

Detailed Solution: Question 15

At lowest point Q: speed is maximum → KE is maximum. Height h = 0 → PE = mgh = 0 (minimum). At extreme positions P and R: v = 0 → KE = 0, height is maximum → PE is maximum. KE + PE = constant at all positions.

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About this Class 9 Practice Test

This test contains 15 MCQs on Work,Energy, and Simple Machines - 1 with detailed solutions for each question. It is part of the Science Class 9 New NCERT 2026-27 (New Syllabus) course on EduRev, designed to align with the current Class 9 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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