Question

A body of mass $5\text{ kg}$ is projected vertically upwards from a point $X$ from the ground with an initial speed of $20\text{ ms}^{-1}$. It rises to a point $Y$ where its kinetic energy is reduced to $400\text{ J}$. If the body experiences a constant air resistance of $10\text{ N}$ throughout its motion, then the vertical distance between $X$ and $Y$ is (Acceleration due to gravity $=10\text{ ms}^{-2}$)

• A. $5\text{ m}$
• B. $10\text{ m}$
• C. $8\text{ m}$
• D. $16\text{ m}$

Hint:

When a body moves upward through resistive air, both gravity and friction work together to drain its kinetic energy: $W = -(mg + F_{\text{air}})h$.

Answer 1

The Correct Option is B

Step 1: Use the work-energy idea.
The change in kinetic energy equals the total work done by all forces. So $W_{\text{net}} = KE_{f} - KE_{i}$.

Step 2: Find the starting kinetic energy.
\[ KE_{i} = \tfrac{1}{2}mv^{2} = \tfrac{1}{2}(5)(20)^{2} = 1000 \text{ J} \]
Step 3: Note the final kinetic energy.
It is given as $KE_{f} = 400$ J. So the change is $\Delta KE = 400 - 1000 = -600$ J.

Step 4: List the forces that slow it down.
Going up, both gravity and air drag push down. Gravity is $mg = 5\times 10 = 50$ N and air drag is $10$ N.

Step 5: Write the work done over height $h$.
Both forces oppose the motion, so \[ W_{\text{net}} = -(50 + 10)h = -60h \]
Step 6: Set the two equal and solve.
\[ -60h = -600 \quad\Rightarrow\quad h = 10 \text{ m} \]So the distance between X and Y is $10$ m, which is option 2.
\[ \boxed{h = 10 \text{ m}} \]