Question

A small drop of mass \(1\,g\) starts falling from rest from a height of \(1\,km\). When it reaches the ground with speed \(5\,m/s\), magnitude of work done by resistance force is \(x\times10^{-3}\,J\). Find \(x\).

• A. \(845\)
• B. \(247.5\)
• C. \(987.5\)
• D. None

Answer 1

The Correct Option is C

Concept: Using work–energy theorem: \[ W_g + W_{res} = \Delta K \] where \[ W_g = mgh \] \[ \Delta K = \frac{1}{2}mv^2 \] Step 1: Write work–energy equation.} \[ mgh + W_{res} = \frac{1}{2}mv^2 \]
Step 2: Solve for work done by resistance.} \[ W_{res} = \frac{1}{2}mv^2 - mgh \]
Step 3: Substitute given values.} \[ m = 1\,g = 10^{-3}\,kg \] \[ h = 1\,km = 1000\,m \] \[ v = 5\,m/s \] \[ W_{res} = 10^{-3}\left(\frac{25}{2} - 10^4\right) \] \[ |W_{res}| = 987.5 \times 10^{-3} J \] Thus \[ x = 987.5 \] Final Result \[ x = 987.5 \]