A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine, so that it rises to a m

Q: Why don't we calculate the launch velocity first?

You certainly can! You could use $v^2 = 2gh$ to find the launch velocity, and then equate the kinetic energy $\frac{1}{2}mv^2$ to the work done $F \cdot d$. However, equating work directly to the final potential energy is faster and elegantly bypasses the intermediate step.

Q: Does it matter that the force is horizontal but the ball goes vertical?

In an ideal machine, the mechanism perfectly redirects the energy. The work done by the input force is fully converted into the ball's mechanical energy, regardless of the direction changes inside the machine.

Q: Why is the initial kinetic energy considered zero?

The ball starts from rest inside the machine before the part begins pushing it. The work done by the machine builds up its kinetic energy from zero.

Q: What happened to the work done by gravity while the ball was being pushed?

The horizontal push implies no vertical displacement during the push phase, so gravity does zero work during the $0.2\text{ m}$ horizontal movement.

Q: Can we use Newton's laws instead of energy?

Yes, but it's much more complicated. You'd have to find the acceleration, then the launch velocity, and then use kinematics for the vertical motion. The Work-Energy theorem is a powerful shortcut!

Question

A cricket ball of mass $0.15	ext{ kg}$ is thrown vertically up by a bowling machine, so that it rises to a maximum height of $20	ext{ m}$ after leaving the machine. If the part pushing the ball applies a constant force $F$ on the ball and moves horizontally a distance of $0.2	ext{ m}$, while launching the ball, the value of $F$ (in N) is (Take, $g = 10	ext{ ms}^{-2}$) ........... .

Sigma Percentile · Accepted Answer

### The Setup: A Horizontal Push for a Vertical Launch

Imagine you are standing next to a cricket bowling machine. Inside its casing, a mechanical arm pushes horizontally with a constant force $F$ over a short distance of $0.2	ext{ m}$. This horizontal work is magically redirected by the machine's internal mechanism to launch the ball straight up into the air!

We are tasked with finding the exact magnitude of this force $F$. To do this, we don't need to worry about the complex gears or springs inside the machine. We just need to look at the energy.

### The Master Equation: Work-Energy Theorem

The Work-Energy Theorem is our ultimate tool here. It states that the work done by the non-conservative force of the machine equals the total change in the ball's mechanical energy.

$$W_{	ext{machine}} = \Delta E = E_f - E_i$$

The ball starts from rest inside the machine, so its initial kinetic energy is zero. It is launched upwards and eventually comes to a momentary halt at its maximum height of $20	ext{ m}$. At this peak, its kinetic energy is again zero, and all the energy is stored as gravitational potential energy.

Therefore, the work done by the machine simply equals the final gravitational potential energy of the ball:

$$F \cdot d = mgh$$

### Crunching the Numbers

Let's substitute the known values into our master equation. The distance $d$ is $0.2	ext{ m}$, the mass of the ball $m$ is $0.15	ext{ kg}$, the acceleration due to gravity $g$ is $10	ext{ ms}^{-2}$, and the maximum height $h$ is $20	ext{ m}$.

$$F 	imes 0.2 = 0.15 	imes 10 	imes 20$$

First, let's calculate the total potential energy on the right side. Multiplying $0.15$ by $10$ gives $1.5$. Multiplying $1.5$ by $20$ gives us exactly $30	ext{ Joules}$ of energy.

$$F 	imes 0.2 = 30$$

### The Final Strike

Now, to isolate the force $F$, we simply divide the total energy by the distance over which the force was applied.

$$F = \frac{30}{0.2}$$

Dividing $30$ by $0.2$ is mathematically equivalent to dividing $300$ by $2$, which yields our final answer:

**$$F = 150	ext{ N}$$**

This means the machine applies a constant force of $150	ext{ Newtons}$ to achieve that impressive $20	ext{ m}$ vertical launch!

Enter Numerical Value:

Visualized Solution

Visualizing the Energy Transfer

Applying Work-Energy Theorem

Substituting the Values

Calculating Potential Energy

Solving for Force

The Way Forward

The Sigma Insight: Work Done by Forces

The Setup

The Master Equation

Crunching the Numbers

The Final Strike

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