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 need to determine the magnitude and direction of this minimum force, and then calculate its horizontal component.

1. Determine the Minimum Force to Drag the Block: 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)
• Frictional force: $f = \mu N$ (opposing motion, i.e., horizontal, backwards)

For vertical equilibrium: $N + F\sin\theta = mg$ $N = mg - F\sin\theta$ (Equation 1)

For horizontal motion (just about to move or moving at constant velocity): $F\cos\theta = f$ $F\cos\theta = \mu N$ (Equation 2)

Substitute Equation 1 into Equation 2: $F\cos\theta = \mu (mg - F\sin\theta)$ $F\cos\theta = \mu mg - \mu F\sin\theta$ $F\cos\theta + \mu F\sin\theta = \mu mg$ $F(\cos\theta + \mu\sin\theta) = \mu mg$ $F = \frac{\mu mg}{\cos\theta + \mu\sin\theta}$

To find the minimum force ($F_{min}$), the denominator $(\cos\theta + \mu\sin\theta)$ must be maximized. This expression is of the form $A\cos\theta + B\sin\theta$, which has a maximum value of $\sqrt{A^2 + B^2}$ when $\tan\theta = B/A$. Here, $A=1$ and $B=\mu$. So, the maximum value of $(\cos\theta + \mu\sin\theta)$ is $\sqrt{1^2 + \mu^2} = \sqrt{1+\mu^2}$. This occurs when $\tan\theta = \mu$.

Thus, the minimum force is: $F_{min} = \frac{\mu mg}{\sqrt{1+\mu^2}}$

2. Calculate the Magnitude of the Minimum Force: Given: Mass $m = \frac{1}{3}$ kg Friction coefficient $\mu = 0.75 = \frac{3}{4}$ Let's use $g = 10 \text{ m/s}^2$ for calculation simplicity, which is a common approximation in such problems if not specified.

First, calculate $\sqrt{1+\mu^2}$: $\sqrt{1+\mu^2} = \sqrt{1 + (0.75)^2} = \sqrt{1 + (3/4)^2} = \sqrt{1 + 9/16} = \sqrt{25/16} = \frac{5}{4}$

Now, calculate $F_{min}$: $F_{min} = \frac{(3/4) \times (1/3) \times 10}{5/4}$ $F_{min} = \frac{(1/4) \times 10}{5/4}$ $F_{min} = \frac{10/4}{5/4} = \frac{10}{5} = 2 \text{ N}$

3. Determine the Horizontal Component of the Minimum Force: The minimum force $F_{min}$ acts at an angle $\theta$ such that $\tan\theta = \mu = 3/4$. From a right triangle with opposite side 3 and adjacent side 4, the hypotenuse is 5. So, $\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4}{5}$.

The horizontal component of the minimum force ($F_{min,x}$) is: $F_{min,x} = F_{min} \cos\theta = 2 \text{ N} \times \frac{4}{5} = \frac{8}{5} = 1.6 \text{ N}$

4. Calculate the Work Done (W): Work done $W = \text{Force} \times \text{Distance}$ (since the force is constant and in the direction of displacement). Distance $d = 5$ m. $W = F_{min,x} \times d$ $W = 1.6 \text{ N} \times 5 \text{ m}$ $W = 8 \text{ J}$

The value of W is 8 joule.