Passage: A small block of mass 1 kg is released from rest at the top of a rough track. The track is

Q: Why is the vertical drop $R \sin 30^\circ$ and not $R \cos 30^\circ$?

The angle $30^\circ$ is given with respect to the horizontal line passing through the center of the arc. The vertical drop is the opposite side to this angle in the right triangle formed, which corresponds to the sine function.

Q: Why do we subtract the work done by friction?

Friction is a non-conservative force that acts opposite to the direction of motion. It removes mechanical energy from the system, which is why its work is considered negative in the energy conservation equation.

Q: Can we use kinematics equations like $v^2 = u^2 + 2as$ here?

No, because the acceleration is not constant. The block is moving along a curved path, and the component of gravity along the path changes continuously. The Work-Energy theorem is the best approach here.

Q: Does the mass of the block affect the final speed?

In this specific problem, the work done by friction is given as an absolute value ($150\text{ J}$). If we change the mass, the kinetic energy and potential energy terms will change, but the friction loss remains $150\text{ J}$, so the final speed would indeed be affected.

Q: What if the track was frictionless?

If the track were frictionless, the entire loss in potential energy would convert into kinetic energy. The equation would simply be $mgh = \frac{1}{2}mv^2$, resulting in a higher final speed.

Question

Passage:
A small block of mass $1	ext{ kg}$ is released from rest at the top of a rough track. The track is a circular arc of radius $40	ext{ m}$. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point $Q$, as shown in the figure, is $150	ext{ J}$. (Take the acceleration due to gravity, $g = 10	ext{ ms}^{-2}$)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the top of a massive, curved water slide. You release a block from rest at the very top, point $P$. The track is a perfect quarter circle with a radius of $40	ext{ m}$. As the block slides down to point $Q$, it doesn't just fall freely; it grinds against the rough surface of the track.

This friction steals some of the block's energy. We are told that the work done to overcome this friction is exactly $150	ext{ J}$. Our mission is to find out how fast the block is moving when it reaches point $Q$.

### The Geometry of the Drop

Before we can talk about energy, we need to understand the geometry of the track. How far has the block actually fallen vertically?

Look at the diagram. The radius line to point $Q$ makes an angle of $30^\circ$ with the horizontal line passing through the center of the arc. The vertical drop $h$ from the top of the track to point $Q$ forms the opposite side of a right-angled triangle.

Using basic trigonometry, we can express this height as:
$$h = R \sin 30^\circ$$

Substituting the given radius $R = 40	ext{ m}$ and the value of $\sin 30^\circ = \frac{1}{2}$, we get:
$$h = 40 	imes \frac{1}{2} = 20	ext{ m}$$

So, the block has descended a vertical distance of $20	ext{ m}$.

### The Master Equation: Work-Energy Theorem

Now, let's bring in the heavy artillery: the Work-Energy Theorem. This powerful principle states that the net work done on an object equals its change in kinetic energy.

In our scenario, gravity is doing positive work by pulling the block down, while friction is doing negative work by resisting the motion. We can write the energy conservation equation as:
$$mgh - W_{	ext{friction}} = \frac{1}{2}mv^2$$

Here, $mgh$ is the initial potential energy that gets converted, $W_{	ext{friction}}$ is the energy lost to heat, and $\frac{1}{2}mv^2$ is the final kinetic energy.

### Final Calculation

Let's plug in the numbers. We know the mass $m = 1	ext{ kg}$, the acceleration due to gravity $g = 10	ext{ ms}^{-2}$, the height $h = 20	ext{ m}$, and the work done against friction is $150	ext{ J}$.

Substituting these into our master equation:
$$1 	imes 10 	imes 20 - 150 = \frac{1}{2} 	imes 1 	imes v^2$$

Simplifying the terms:
$$200 - 150 = 0.5 v^2$$

$$50 = 0.5 v^2$$

Dividing both sides by $0.5$:
$$v^2 = 100$$

Taking the square root gives us the final speed:
$$v = 10	ext{ ms}^{-1}$$

**The block is moving at $10	ext{ ms}^{-1}$ when it reaches point $Q$.**

Animated Solution for Physics - Work, Energy, and Power: Comprehension Passage

Visualized Solution

Visualizing the Track

Work-Energy Theorem

Calculating Vertical Drop $h$

Evaluating $h$

Setting up the Energy Equation

Simplifying the Equation

Solving for $v^{2}$

Final Velocity

The Sigma Insight: Work Done by Forces

Analyzing the Setup

The Geometry of the Drop

The Master Equation

Final Calculation

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