A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.

Q: Why is the change in kinetic energy zero? Doesn't the ball move?

Yes, the ball moves, but the Work-Energy Theorem looks at the *initial* and *final* states. The ball starts from rest in your hand ($v=0$) and momentarily stops at its maximum height ($v=0$). Since both velocities are zero, the overall change in kinetic energy is zero.

Q: Why does gravity do work over 2.2m and not just 2m?

Gravity is acting on the ball from the very first millisecond you start pushing it. It doesn't wait for the ball to leave your hand! Therefore, gravity does negative work over the 0.2m push AND the 2m free flight, totaling 2.2m.

Q: Can I solve this using $F=ma$ and kinematics instead?

Absolutely! You can use $v^2 = u^2 + 2as$ for the free flight to find the launch velocity, then use it again for the pushing phase to find the acceleration. Finally, $F - mg = ma$ will give you the same answer. The Work-Energy method is just a faster shortcut!

Q: Why is the work done by gravity negative?

Work is the dot product of force and displacement ($W = \vec{F} \cdot \vec{d}$). Gravity pulls downwards, but the ball is moving upwards. Since the force and displacement are in opposite directions (180 degrees apart), the work done is negative.

Q: What happens to the energy after the hand releases the ball?

During the push, the hand does positive work, giving the ball kinetic energy. Once it leaves the hand, gravity does negative work, slowly draining that kinetic energy and converting it into gravitational potential energy until the ball stops at the top.

Question

A ball of mass $0.2	ext{ kg}$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2	ext{ m}$ while applying the force and the ball goes upto $2	ext{ m}$ height further, find the magnitude of the force. Consider $g = 10	ext{ m/s}^2$.

Options:

(A) $4	ext{ N}$

(B) $16	ext{ N}$

(C) $20	ext{ N}$

(D) $22	ext{ N}$

Sigma Percentile · Accepted Answer

Imagine you are holding a ball in your hand. You decide to throw it straight up into the air. To do this, your hand pushes upwards against the ball for a short distance of $0.2	ext{ m}$. After the ball leaves your hand, it continues to fly upwards for another $2	ext{ m}$ before gravity finally brings it to a momentary halt at its peak.

Our goal is to find the exact magnitude of the force your hand applied during that initial $0.2	ext{ m}$ push. While we could use the equations of kinematics and Newton's Second Law ($F=ma$) to solve this, there is a much more elegant and direct approach: **The Work-Energy Theorem**.

### The Elegance of the Work-Energy Theorem

The Work-Energy Theorem states that the net work done on an object by all forces equals its change in kinetic energy:

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

Why is this so powerful here? Because it allows us to completely ignore the intermediate velocity of the ball as it leaves your hand. We only need to look at the very beginning and the very end of the journey.

### Breaking Down the Journey

Let's analyze the kinetic energy at the start and the end:
- **Initial State:** The ball starts from rest in your hand. So, initial velocity $u = 0$, which means initial kinetic energy $K_i = 0$.
- **Final State:** The ball reaches its maximum height and momentarily stops. So, final velocity $v = 0$, which means final kinetic energy $K_f = 0$.

Therefore, the total change in kinetic energy over the entire journey is zero:

$$\Delta K = 0$$

### Calculating the Work Done

Now, let's identify the forces doing work on the ball. There are only two:
1. **Gravity ($mg$):** Gravity pulls downwards on the ball from the very first millisecond you start pushing it, all the way until it reaches the top. The total distance the ball travels upwards is $0.2	ext{ m} + 2	ext{ m} = 2.2	ext{ m}$. Since gravity acts downwards and the ball moves upwards, the work done by gravity is negative.
   $$W_g = -mg 	imes h_{	ext{total}} = -(0.2)(10)(2.2) = -4.4	ext{ J}$$

2. **Applied Force ($F$):** Your hand pushes upwards, and the ball moves upwards. This force is only applied over the first $0.2	ext{ m}$. Since the force and displacement are in the same direction, the work done is positive.
   $$W_F = F 	imes d_{	ext{hand}} = F(0.2)$$

### The Final Computation

According to the Work-Energy Theorem, the sum of the work done by all forces must equal the change in kinetic energy:

$$W_g + W_F = \Delta K$$

Substituting our calculated values:

$$-4.4 + 0.2F = 0$$

Now, we simply solve for $F$:

$$0.2F = 4.4$$

$$F = \frac{4.4}{0.2} = 22	ext{ N}$$

And there we have it! By looking at the energy of the system as a whole, we bypassed the need to calculate accelerations and velocities, arriving at the answer of **$22	ext{ N}$** swiftly and elegantly.

Animated Solution for Physics - Work, Energy, and Power: A ball of mass $0.2 kg$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2 m$ while applying the force and the ball goes upto $2 m$ height further, find the magnitude of the force. Consider $g = 10 m/s^{2}$ .

Select Answer:

Visualized Solution

\text{Visualizing the Journey}

\text{The Work-Energy Theorem}

\text{Change in Kinetic Energy}

\text{Work Done by Forces}

\text{Setting up the Equation}

\text{Solving for } F

\text{Final Force Magnitude}

\text{Alternative Approach}

The Sigma Insight: Work Done by Forces

The Elegance of the Work-Energy Theorem

Breaking Down the Journey

Calculating the Work Done

The Final Computation

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