A body of mass m dropped from a height h reaches the ground with a speed of 0.8\sqrt{gh}. The value

Q: Why is the initial kinetic energy taken as zero?

The problem states that the body is 'dropped'. In physics, 'dropped' implies that the object is released from rest, meaning its initial velocity is zero. Therefore, its initial kinetic energy is also zero.

Q: Why is the work done by air friction negative?

Work is defined as the dot product of force and displacement ($W = \vec{F} \cdot \vec{d}$). Air friction always acts opposite to the direction of motion. Since the displacement is downwards and the friction is upwards, the angle between them is $180^\circ$. $\cos(180^\circ) = -1$, resulting in negative work.

Q: Can we use the equations of kinematics (like $v^2 = u^2 + 2as$) to solve this?

No, you cannot reliably use standard kinematic equations here. Air friction is typically dependent on velocity, meaning the net force (and thus acceleration) is not constant during the fall. The Work-Energy Theorem is the safest and most direct approach because it doesn't require constant acceleration.

Q: What would the final velocity be if there was no air friction?

If there were no air friction, mechanical energy would be perfectly conserved. The potential energy $mgh$ would completely convert into kinetic energy $\frac{1}{2}mv^2$, giving a final velocity of $v = \sqrt{2gh}$.

Q: Does the mass of the body affect the work done by air friction in this specific problem?

Yes, based on our final derived expression ($-0.68mgh$), the work done by air friction is directly proportional to the mass $m$ of the body. A heavier body falling under these exact same kinematic conditions would experience more work done by air friction.

Question

A body of mass $m$ dropped from a height $h$ reaches the ground with a speed of $0.8\sqrt{gh}$. The value of workdone by the air-friction is

Options:

(A) $-0.68mgh$

(B) $mgh$

(C) $1.64mgh$

(D) $0.64mgh$

Sigma Percentile · Accepted Answer

### The Physics of a Falling Body

Imagine a ball of mass $m$ dropped from a height $h$. If we lived in a perfect vacuum, the only force acting on the ball would be gravity. It would accelerate downwards at $g$, and by the time it hit the ground, its velocity would be exactly $\sqrt{2gh}$.

However, the real world is messy. As the ball falls through the atmosphere, it collides with countless air molecules. These collisions create a resistive force known as **air friction** or drag. This force pushes upwards, directly opposing the downward motion of the ball. Because of this resistance, the ball doesn't reach the theoretical maximum speed. In our problem, it only reaches a speed of $0.8\sqrt{gh}$.

### The Master Equation: Work-Energy Theorem

To figure out exactly how much energy the air friction stole from the ball, we turn to one of the most powerful tools in classical mechanics: the **Work-Energy Theorem**.

The theorem states that the net work done by all forces acting on an object equals the change in its kinetic energy:

$$W_{	ext{net}} = \Delta K$$

Let's break down the forces. We have gravity pulling down and air friction pushing up. Therefore, the net work is the sum of the work done by gravity ($W_{	ext{gravity}}$) and the work done by air friction ($W_{	ext{air-friction}}$).

$$W_{	ext{gravity}} + W_{	ext{air-friction}} = K_f - K_i$$

### Setting Up the Variables

Since the ball is dropped from rest, its initial velocity $u = 0$, which means its initial kinetic energy $K_i = 0$.

As the ball falls through a height $h$, gravity does positive work because the force and displacement are in the same direction.

$$W_{	ext{gravity}} = mgh$$

The final kinetic energy $K_f$ is calculated using the final velocity $v_f = 0.8\sqrt{gh}$:

$$K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}m(0.8\sqrt{gh})^2$$

### The Final Calculation

Now, we carefully square the final velocity term. The square of $0.8$ is $0.64$, and the square root of $gh$ simply becomes $gh$.

$$K_f = \frac{1}{2}m(0.64gh) = 0.32mgh$$

Substituting everything back into our master equation:

$$mgh + W_{	ext{air-friction}} = 0.32mgh - 0$$

To isolate the work done by air friction, we subtract $mgh$ from both sides:

$$W_{	ext{air-friction}} = 0.32mgh - mgh$$

$$W_{	ext{air-friction}} = -0.68mgh$$

**The negative sign is crucial here.** It mathematically confirms our physical intuition: air friction acts in the opposite direction of the displacement, thereby doing negative work and removing mechanical energy from the system. The correct option is **(a)**.

Animated Solution for Physics - Work, Energy, and Power: A body of mass $m$ dropped from a height $h$ reaches the ground with a speed of $0.8 g h$ . The value of workdone by the air-friction is

Select Answer:

Visualized Solution

Visualizing the Fall

Work-Energy Theorem

Substituting the Values

Calculating Final Kinetic Energy

Simplifying the Equation

Solving for Air Friction Work

Conclusion

The Sigma Insight: Work Done by Forces

The Physics of a Falling Body

The Master Equation

Setting Up the Variables

The Final Calculation

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