Question

A body is allowed to fall freely under gravity from a height of 15 m from the ground. At a point in its path, if the kinetic energy of the body is 200% more than its potential energy, then the velocity of the body at that point is (Acceleration due to gravity = 10 $ms^{-2}$)

• A. 6 $ms^{-1}$
• B. 20 $ms^{-1}$
• C. 10 $ms^{-1}$
• D. 15 $ms^{-1}$

Hint:

Interpretation of percentages: - "is 200% of X" $\rightarrow 2X$ - "is 200% {more than} X" $\rightarrow X + 2X = 3X$ Always read the wording carefully.

Answer 1

The Correct Option is D

Step 1: Energy Analysis:
Let $PE$ be the Potential Energy and $KE$ be the Kinetic Energy at that specific point. Condition given: "KE is 200% more than PE". \[ KE = PE + \frac{200}{100}PE \] \[ KE = PE + 2PE = 3PE \]
Step 2: Conservation of Mechanical Energy:
The total mechanical energy ($TE$) is conserved and is equal to the initial potential energy at height $H = 15$ m. \[ TE = mgH \] Also, at the specific point: \[ TE = KE + PE \] Substitute $KE = 3PE$: \[ TE = 3PE + PE = 4PE \] From this, $PE = \frac{TE}{4} = \frac{mgH}{4}$. Consequently, $KE = 3PE = \frac{3mgH}{4}$.
Step 3: Calculating Velocity:
We know $KE = \frac{1}{2}mv^2$. Equating the expressions for KE: \[ \frac{1}{2}mv^2 = \frac{3}{4}mgH \] Cancel mass $m$ and solve for $v$: \[ v^2 = \frac{3}{2}gH \]
Step 4: Substitution:
Given $g = 10 \, ms^{-2}$ and $H = 15$ m. \[ v^2 = \frac{3}{2} \times 10 \times 15 \] \[ v^2 = 1.5 \times 150 = 225 \] \[ v = \sqrt{225} = 15 \, ms^{-1} \]
Step 5: Final Answer:
The velocity of the body is 15 $ms^{-1}$.