A body of mass 2 kg is being dragged with a uniform velocity of 2 m/s on a rough horizontal plane. T

Q: Wait, why is the net force zero if the block is moving?

According to Newton's First Law, an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Since the velocity is uniform (constant), the acceleration is zero, meaning the net force must be zero. The force pulling the block perfectly balances the friction.

Q: Why do we use $mg$ for the normal force?

Because the block is moving on a flat, horizontal surface and there are no other vertical forces acting on it. The vertical forces must balance out (equilibrium), so the upward normal force $N$ exactly equals the downward gravitational force $mg$.

Q: What exactly is $J = 4.2\text{ J/cal}$?

It is the mechanical equivalent of heat. It acts as a conversion factor between two different units of energy: Joules (typically used for mechanical work) and calories (typically used for heat). It tells us that $4.2\text{ Joules}$ of mechanical work will produce $1\text{ calorie}$ of heat.

Q: Does all the work done by friction turn into heat?

In this idealized macroscopic physics model, yes. We assume that $100\%$ of the mechanical energy dissipated by kinetic friction is converted into thermal energy (heat) that warms up the block and the surface.

Q: Why doesn't the kinetic energy of the block change?

Because the velocity is constant! The Work-Energy Theorem states that the net work done on an object equals its change in kinetic energy ($W_{\text{net}} = \Delta K$). Since $\Delta K = 0$, the net work is zero. The positive work done by the pulling force is exactly canceled out by the negative work done by friction.

Question

A body of mass $2	ext{ kg}$ is being dragged with a uniform velocity of $2	ext{ m/s}$ on a rough horizontal plane. The coefficient of friction between the body and the surface is $0.20$, $J = 4.2	ext{ J/cal}$ and $g = 9.8	ext{ m/s}^2$. Calculate the amount of heat generated in $5	ext{ s}$.

Sigma Percentile · Accepted Answer

### The Illusion of Effortless Motion

Imagine dragging a heavy $2	ext{ kg}$ block across a rough floor. You are pulling it, and it's moving at a perfectly steady pace of $2	ext{ m/s}$. Because the velocity is uniform, Newton's First Law tells us that the net force on the block is zero. This means whatever force you are applying to pull the block is being exactly canceled out by an invisible, opposing force: kinetic friction.

### Calculating the Drag Distance

Before we can figure out how much energy is lost to friction, we need to know how far the block was dragged. Since the block is not accelerating, we can use the simplest equation of motion: distance equals velocity multiplied by time.

$$s = vt$$

Plugging in our given values of $v = 2	ext{ m/s}$ and $t = 5	ext{ s}$, we find that the block travels a total distance of $10	ext{ meters}$. That is $10	ext{ meters}$ of continuous grinding against the rough floor!

### The Invisible Firestarter: Friction

As the block slides, the microscopic irregularities of the block and the floor crash into each other. This kinetic friction is what generates heat. The magnitude of this frictional force on a flat horizontal surface is proportional to the normal force, which in this case is simply the weight of the block.

$$f_k = \mu N = \mu mg$$

The work done against this frictional force is the mechanical energy that gets dissipated as heat. Work is defined as force times displacement.

$$W = f_k 	imes s = \mu mg s$$

Substituting our known values ($\mu = 0.20$, $m = 2	ext{ kg}$, $g = 9.8	ext{ m/s}^2$, and $s = 10	ext{ m}$), we can calculate the total work done.

$$W = 0.20 	imes 2 	imes 9.8 	imes 10 = 39.2	ext{ J}$$

### The Mechanical Equivalent of Heat

We have found that $39.2	ext{ Joules}$ of mechanical work was done. However, the problem provides a specific constant: $J = 4.2	ext{ J/cal}$. This is known as the mechanical equivalent of heat, a historic conversion factor discovered by James Prescott Joule, which links mechanical energy (Joules) to thermal energy (calories).

To find the heat $Q$ generated in calories, we simply divide our work in Joules by this constant $J$.

$$Q = \frac{W}{J} = \frac{39.2}{4.2}$$

Carrying out this final division, we arrive at our answer.

**$Q = 9.33	ext{ cal}$**

And there we have it! The mechanical effort of dragging the block has been transformed into exactly $9.33	ext{ calories}$ of thermal energy, warming up the block and the floor.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Kinematics of Uniform Motion

Substituting Values for Distance

Calculating Distance

Identifying the Forces

Work Done by Friction

Substituting Values for Work

Calculating Work Done

Mechanical Equivalent of Heat

Substituting Values for Heat

Final Calculation

The Sigma Insight: Work Done by Forces

The Illusion of Effortless Motion

Calculating the Drag Distance

The Invisible Firestarter

The Mechanical Equivalent of Heat

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