Question

A body is projected vertically upwards with a velocity of \( 20~\text{ms}^{-1} \). If the potential energy of the body at a height of \( 5~\text{m} \) from the ground is \( 100~\text{J} \), then the kinetic energy of the body at a height of \( 10~\text{m} \) from the ground is
\textit{(Acceleration due to gravity \( g = 10~\text{ms}^{-2} \))}

• A. \( 200~\text{J} \)
• B. \( 300~\text{J} \)
• C. \( 150~\text{J} \)
• D. \( 250~\text{J} \)

Hint:

Use conservation of mechanical energy: Total energy remains constant if only conservative forces (like gravity) are acting. Find kinetic energy by subtracting potential energy from total energy.

Answer 1

The Correct Option is A

Step 1: Find total mechanical energy using data at 5 m. \[ \text{Total Energy} = \text{K.E.} + \text{P.E.} \] At 5 m, P.E. is given as \( 100~\text{J} \).
Let total energy be \( E \), then: \[ E = K.E._{5\text{m}} + 100~\text{J} \] Step 2: Use velocity to find total mechanical energy. \[ \text{Initial K.E.} = \frac{1}{2}mv^2 = \frac{1}{2} \times m \times (20)^2 = 200m \] Let mass be \( m \), and use P.E. at 5 m: \[ P.E. = mgh = m \times 10 \times 5 = 50m \Rightarrow 50m = 100 \Rightarrow m = 2~\text{kg} \] So total mechanical energy: \[ E = 200 \times 2 = 400~\text{J} \] Step 3: Find P.E. at 10 m: \[ P.E. = mgh = 2 \times 10 \times 10 = 200~\text{J} \] Step 4: Use conservation of energy to find K.E. at 10 m: \[ K.E. = E - P.E. = 400 - 200 = 200~\text{J} \] % Final Answer \[ \boxed{200~\text{J}} \]