A block of mass $m$ is released from rest at a height $h$ on a smooth inclined plane of inclination $\theta$. The kinetic energy of the block at the bottom of the incline is:

## CORE INFORMATION\n✓ **Answer**: Option C\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Conservation of Mechanical Energy\n\t**Related Topics**: Potential Energy, Kinetic Energy\n\t**Prerequisites**: Understanding of potential and kinetic energy, the concept of conservation of energy.\n## SOLUTION PATHWAY\n**Given Information**:\n\t* Mass of the block: $m$\n\t* Initial height: $h$\n\t* Initial velocity: 0 (released from rest)\n\t* Smooth inclined plane (no friction, no energy loss due to friction)\n\t* Inclination of plane: $\theta$\n**Method & Steps**:\n\t1. **Identify initial and final energies**\n\t\tReasoning: At the top, the block has only potential energy since it is at rest. At the bottom, the block has only kinetic energy since we consider the potential energy reference to be at the bottom of the incline.\n\t\tFormula/Rule: Potential energy (PE) = $mgh$, Kinetic energy (KE) = $\frac{1}{2}mv^2$\n\t2. **Apply the law of conservation of energy**\n\t\tWorking: Since the plane is smooth, there is no loss of energy due to friction. Therefore, the total mechanical energy (PE + KE) at the top must equal the total mechanical energy at the bottom. Initial KE = 0, Initial PE = mgh. Final KE = KE_bottom, Final PE = 0\n\t\tKey Point: $PE_{initial} + KE_{initial} = PE_{final} + KE_{final}$\n\t3. **Solve for final kinetic energy**\n\t\tResult: $mgh + 0 = 0 + KE_{bottom}$. Therefore, $KE_{bottom} = mgh$\n\t\tVerification: The final kinetic energy equals the initial potential energy, which is expected due to conservation of energy.\n## OPTION ANALYSIS\n[A] :x: $mgh\sin\theta$ - Incorrect. This is the potential energy component along the incline, not the final kinetic energy.\n[B] :x: $mgh\cos\theta$ - Incorrect. This is related to the normal force on the incline and not the final kinetic energy.\n[C] ✓: $mgh$ - Correct. The kinetic energy at the bottom is equal to the initial potential energy at the top due to conservation of energy.\n[D] :x: $\frac{1}{2}mgh$ - Incorrect. This would indicate a loss of half the initial potential energy, which is not the case for a smooth incline.\n## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1. Misunderstanding potential energy: Forgetting that potential energy is calculated using the vertical height, not the distance along the incline. Always use the vertical height difference.\n\t\t2. Incorrectly calculating the final kinetic energy: Assuming that some of the initial energy is lost due to friction, when the question states a smooth incline.\n\t**Quick Tips**:\n\t\t1. Conservation of Energy: For a frictionless system, the initial total mechanical energy equals the final total mechanical energy. This simplifies problems substantially.\n\t\t2. Focus on the vertical height: When calculating potential energy, always use the vertical height relative to the reference point.

NEET Physics Work, Energy, and Power Kinetic Energy MCQs

Prepare for NEET Physics Work, Energy, and Power (Kinetic Energy) with MCQs & PYQs on NEET.GUIDE. Access free practice, previous year questions, and expert solutions to understand energy associated with motion.

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Work Done By Constant Force

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A particle of mass $m$ moving with velocity $v$ collides elastically with another particle of mass $2m$ at rest. If the lighter particle is deflected through an angle of $90^\circ$ , the fractional loss in its kinetic energy is:

1/3

1/2

2/3

3/4

Question Tags

📚 Subject:Physics

📑 Chapter:Work, Energy, and Power

🎯Topic:Kinetic Energy

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :3 mins

🔄Ncert Concepts :Kinetic Energy, Momentum, Collision, Elastic Collision

Two identical balls A and B are moving with velocities $v$ and $2v$ respectively, having the same kinetic energy. The ratio of their masses is:

1:1

2:1

1:2

4:1

Question Tags

📚 Subject:Physics

📑 Chapter:Work, Energy, and Power

🎯Topic:Kinetic Energy

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :1 mins

🔄Ncert Concepts :Kinetic Energy, Mass, Velocity

A body of mass $m$ is moving in a circular path of radius $r$ with a constant speed $v$ . The work done by the centripetal force in half a revolution is:

$\pi mv^2$

$2\pi mv^2$

$\frac{mv^2}{r}$

📚 Subject:Physics

📑 Chapter:Work, Energy, and Power

🎯Topic:Kinetic Energy

🎓 Ncert Level:Application

🎯Difficulty :Easy

⚡ Time Required :1 mins

🔄Ncert Concepts :Kinetic Energy, Speed, Calculation

10.

Which of the following has the greatest kinetic energy? A) A 2 kg object moving at 1 m/s B) A 1 kg object moving at 2 m/s C) A 0.5 kg object moving at 4 m/s D) A 4 kg object moving at 0.5 m/s

A) A 2 kg object moving at 1 m/s

B) A 1 kg object moving at 2 m/s

C) A 0.5 kg object moving at 4 m/s

D) A 4 kg object moving at 0.5 m/s