Work Done Calculation Overview

This section explains how to calculate the work done when moving an object across a horizontal surface, considering the effects of friction.

Friction Fundamentals and Physics

Understanding the forces involved is key. We'll look at:

• Work Done ($W$): Defined as force multiplied by distance in the direction of the force. The formula is $W = F \times d$.
• Frictional Force ($F_f$): The force resisting motion. For kinetic friction, it's calculated as $F_f = \mu \times N$, where $\mu$ is the coefficient of friction and $N$ is the normal force.
• Normal Force ($N$): For an object resting on a horizontal plane, the normal force is equal in magnitude to its weight.
• Weight ($W_{body}$): The force due to gravity, calculated as mass times gravitational acceleration ($W_{body} = m \times g$).

Physics Calculation Steps

Here’s a step-by-step breakdown using the provided values:

1. Normal Force Calculation

Since the body is on a horizontal plane, the normal force ($N$) exerted by the plane on the body is equal in magnitude to the body's weight.

Given values:

• Mass ($m$): 100 kg
• Coefficient of Friction ($\mu$): 0.025
• Distance ($d$): 10 metres

The options provided use the unit 'kg metres'. This suggests the calculation might be expected in units of kilogram-force metres (kgf·m). In such a context, the normal force can be considered numerically equal to the mass when expressed in kilogram-force.

Normal Force ($N$) $\approx$ 100 kg-force

(For reference, using standard SI units: $N = m \times g = 100 \, \text{kg} \times 9.8 \, m/s^2 = 980 \, \text{N}$)

2. Frictional Force Calculation

The horizontal force required to move the body at a constant velocity is equal in magnitude to the kinetic frictional force.

Using the formula: $F_f = \mu \times N$

Applying the values in the context suggested by the options:

$F_f = 0.025 \times 100 \, \text{kg-force} = 2.5 \, \text{kg-force}$

(Using standard SI units: $F_f = 0.025 \times 980 \, \text{N} = 24.5 \, \text{N}$)

3. Work Done Calculation

Work Done ($W$) is calculated as the force applied multiplied by the distance moved in the direction of the force.

Formula: $W = F_f \times d$

Calculating the work done using the units consistent with the options:

$W = 2.5 \, \text{kg-force} \times 10 \, \text{metres} = 25 \, \text{kg-force metres}$

The unit 'kg metres' in the options is interpreted as kilogram-force metres.

(Using standard SI units: $W = 24.5 \, \text{N} \times 10 \, \text{m} = 245 \, \text{Joules}$)

(To convert Joules to kg-force metres, we divide by the approximate value of $g$: $W = \frac{245 \, \text{J}}{9.8 \, m/s^2} = 25 \, \text{kg-force metres}$)

Final Result Analysis

The calculation confirms that the work done in moving the body against friction is 25 kg metres.