Question

A body of mass m dropped from a height h reaches the ground with a speed of 0 . 8 √ g h . The value of work done by the air-friction is:

• A. 0.64 mgh
• B. 0.32 mgh
• C. -0.32 mgh
• D. -0.68 mgh

Answer 1

Answer: (D)

-0.68 mgh

Answer 2

The problem can be solved using the Work-Energy Theorem. The Work-Energy Theorem states that the net work done on an object is equal to the change in its kinetic energy.

1. Identify Initial and Final States:

• Initial State: The body is dropped from a height $h$.
  • Initial velocity, $v_i = 0$
  • Initial kinetic energy, $K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (0)^2 = 0$
• Final State: The body reaches the ground (height = 0) with a speed of $0.8 \sqrt{gh}$.
  • Final velocity, $v_f = 0.8 \sqrt{gh}$
  • Final kinetic energy, $K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (0.8 \sqrt{gh})^2 = \frac{1}{2} m (0.64 gh) = 0.32 mgh$

2. Calculate the Change in Kinetic Energy: $\Delta K = K_f - K_i = 0.32 mgh - 0 = 0.32 mgh$

3. Identify Forces and Work Done by Each: The forces acting on the body are:

• Gravity: A conservative force, acting downwards.
  • Work done by gravity, $W_{gravity} = \text{Force} \times \text{displacement} = mg \times h = mgh$ (since both force and displacement are in the same direction).
• Air Friction: A non-conservative (dissipative) force, opposing the motion.
  • Work done by air friction, $W_{air-friction}$ (this is what we need to find).

4. Apply the Work-Energy Theorem: The net work done ($W_{net}$) is the sum of the work done by all forces: $W_{net} = W_{gravity} + W_{air-friction}$

According to the Work-Energy Theorem: $W_{net} = \Delta K$

Therefore, $W_{gravity} + W_{air-friction} = \Delta K$

5. Solve for Work Done by Air Friction: Substitute the values we calculated: $mgh + W_{air-friction} = 0.32 mgh$

$W_{air-friction} = 0.32 mgh - mgh$ $W_{air-friction} = (0.32 - 1) mgh$ $W_{air-friction} = -0.68 mgh$

The negative sign indicates that the work done by air friction is against the direction of motion, as expected for a resistive force.

Therefore, the value of work done by the air-friction is $-0.68 mgh$.