If the ratio of the potential and kinetic energies of a body of mass 3 kg thrown vertically upwards at a height of 40 m is 2:3, then the kinetic energy of the body at a height of 65 m is (Acceleration due to gravity = 10 ms-2)

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q87 – AP-EAMCET 2026

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Home Physics Work, Energy & PowerQuestion #641

If the ratio of the potential and kinetic energies of a body of mass

3 kg

thrown vertically upwards at a height of

40 m

2 : 3

, then the kinetic energy of the body at a height of

65 m

is (Acceleration due to gravity =

10 {ms}^{- 2}

)

Clue 1 of 3: Conservation of Mechanical Energy

For a body moving under gravity, the total mechanical energy (sum of potential and kinetic energy) remains constant throughout its flight, assuming no air resistance.

🧠 Full Solution Path

Step 1: Calculate the Potential Energy at the first height (

h_{1}

).✦ Active

The potential energy ( $P E_{1}$ ) of the body at height $h_{1} = 40 m$ is calculated using the formula $P E = m g h$ , where $m = 3 kg$ and $g = 10 {m/s}^{2}$ .

P E_{1} = m g h_{1}

= (3 kg) (10 {m/s}^{2}) (40 m)

= 1200 J

💡 Teacher's Secret Hint

Remember to use consistent units (SI units in this case) for mass, acceleration due to gravity, and height to ensure correct energy calculations.

Step 2: Determine Kinetic Energy at

h_{1}

and Total Mechanical Energy.○ Expand

Given the ratio of potential energy to kinetic energy at $h_{1}$ is $2 : 3$ , we can find $K E_{1}$ . The total mechanical energy ( $E$ ) is the sum of $P E_{1}$ and $K E_{1}$ . Since only conservative forces (gravity) are acting, the total mechanical energy is conserved throughout the motion.

\frac{P E_{1}}{K E_{1}} = \frac{2}{3}

K E_{1} = \frac{3}{2} P E_{1} = \frac{3}{2} (1200 J) = 1800 J

E = P E_{1} + K E_{1} = 1200 J + 1800 J = 3000 J

💡 Teacher's Secret Hint

The total mechanical energy $E$ will remain constant at any height as long as air resistance is negligible, making it a powerful tool for solving problems at different points in the trajectory.

Step 3: Calculate the Potential Energy at the second height (

h_{2}

).○ Expand

Now, calculate the potential energy ( $P E_{2}$ ) of the body at the new height $h_{2} = 65 m$ , using the same formula.

P E_{2} = m g h_{2}

= (3 kg) (10 {m/s}^{2}) (65 m)

= 1950 J

Step 4: Determine Kinetic Energy at

h_{2}

using conservation of energy.○ Expand

Using the principle of conservation of mechanical energy, the total energy $E$ at height $h_{2}$ must be equal to the total energy calculated in Step 2. We can then find the kinetic energy $K E_{2}$ at this height.

E = P E_{2} + K E_{2}

3000 J = 1950 J + K E_{2}

K E_{2} = 3000 J - 1950 J

K E_{2} = 1050 J

💡 Teacher's Secret Hint

Always double-check your arithmetic when subtracting or adding energy values to avoid simple calculation errors.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 87 – AP-EAMCET 2026

Clue 1 of 3: Conservation of Mechanical Energy

🧠 Full Solution Path