Question

A block of mass $m = \frac{1}{3}$ kg is kept on a rough horizontal plane. Friction coefficient is $\mu = 0.75.$ The work done by minimum force required to drag the block along the plane by a distance 5 m, is W joule, then find the value of W.

Answer 1

8

Answer 2

To find the work done by the minimum force required to drag the block, we first need to determine the magnitude of this minimum force and the angle at which it acts.

1. Analyze the Forces: Let the applied force be $F$ acting at an angle $\theta$ above the horizontal. The forces acting on the block are:

• Gravitational force: $mg$ (downwards)
• Normal force: $N$ (upwards)
• Applied force: $F$ (at angle $\theta$ to horizontal)
• Kinetic friction force: $f_k$ (opposite to the direction of motion)

2. Set up Equations of Equilibrium/Motion: Since the block is dragged along a horizontal plane, there is no vertical acceleration.

• Vertical equilibrium: $N + F \sin\theta = mg \implies N = mg - F \sin\theta$
• Horizontal motion: For the block to be dragged, the horizontal component of the applied force must overcome the friction force. For minimum force, it just balances friction: $F \cos\theta = f_k$

3. Express Friction Force: The kinetic friction force is given by $f_k = \mu N$. Substituting the expression for $N$: $f_k = \mu (mg - F \sin\theta)$

4. Solve for the Applied Force F: Substitute the expression for $f_k$ into the horizontal motion equation: $F \cos\theta = \mu (mg - F \sin\theta)$ $F \cos\theta = \mu mg - \mu F \sin\theta$ $F \cos\theta + \mu F \sin\theta = \mu mg$ $F (\cos\theta + \mu \sin\theta) = \mu mg$ $F = \frac{\mu mg}{\cos\theta + \mu \sin\theta}$

5. Find the Angle for Minimum Force: To minimize $F$, the denominator $(\cos\theta + \mu \sin\theta)$ must be maximized. We can express $\cos\theta + \mu \sin\theta$ in the form $R \cos(\theta - \alpha)$, where $R = \sqrt{1^2 + \mu^2}$ and $\tan\alpha = \mu$. The maximum value of $R \cos(\theta - \alpha)$ is $R = \sqrt{1 + \mu^2}$, which occurs when $\theta = \alpha$. Therefore, the minimum force $F_{min}$ occurs when $\tan\theta = \mu$. The minimum force is $F_{min} = \frac{\mu mg}{\sqrt{1 + \mu^2}}$.

6. Calculate the Work Done (W): The work done by the force $F$ is $W = (F \cos\theta) \times d$, where $d$ is the distance. At the angle for minimum force, $\tan\theta = \mu$. We can construct a right triangle with opposite side $\mu$, adjacent side 1, and hypotenuse $\sqrt{1 + \mu^2}$. From this, $\cos\theta = \frac{1}{\sqrt{1 + \mu^2}}$.

Substitute $F_{min}$ and $\cos\theta$ into the work done formula: $W = \left(\frac{\mu mg}{\sqrt{1 + \mu^2}}\right) \left(\frac{1}{\sqrt{1 + \mu^2}}\right) \times d$ $W = \frac{\mu mg}{1 + \mu^2} \times d$

7. Substitute Given Values: Given:

• Mass $m = \frac{1}{3}$ kg
• Friction coefficient $\mu = 0.75 = \frac{3}{4}$
• Distance $d = 5$ m
• Use $g = 10$ m/s$^2$ (standard value in absence of specific instruction)

Calculate $1 + \mu^2$: $1 + \mu^2 = 1 + (0.75)^2 = 1 + \left(\frac{3}{4}\right)^2 = 1 + \frac{9}{16} = \frac{16+9}{16} = \frac{25}{16}$

Now, substitute all values into the work done formula: $W = \frac{\left(\frac{3}{4}\right) \times \left(\frac{1}{3}\right) \times 10}{\frac{25}{16}} \times 5$ $W = \frac{\frac{1}{4} \times 10}{\frac{25}{16}} \times 5$ $W = \frac{\frac{10}{4}}{\frac{25}{16}} \times 5$ $W = \frac{10}{4} \times \frac{16}{25} \times 5$ $W = \frac{5}{2} \times \frac{16}{25} \times 5$ $W = \frac{5 \times 16 \times 5}{2 \times 25}$ $W = \frac{25 \times 16}{50}$ $W = \frac{16}{2}$ $W = 8$ J

The value of W is 8 joule.