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Chapter 3: Problem 374

Two blocks of mass \(8 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are connected a heavy string Placed on rough horizontal Plane, The \(4 \mathrm{~kg}\) block is Pulled with a constant force \(\mathrm{F}\). The co-efficient of friction between the blocks and the ground is \(0.5\), what is the value of \(F\), So that the tension in the spring is constant throughout during the motion of the blocks? \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) (A) \(40 \mathrm{~N}\) (B) \(60 \mathrm{~N}\) (C) \(50 \mathrm{~N}\) (D) \(30 \mathrm{~N}\)

Short Answer

Expert verified
The force required to keep the tension constant throughout the motion is \(40 \mathrm{~N}\). The correct answer is (A) \(40 \mathrm{~N}\).

Step by step solution

01

Calculate the friction force on each block

We know that the friction force is given by: \(f = \mu N\), where \(f\) is the friction force, \(\mu\) is the coefficient of friction, and \(N\) is the normal force on the object. For both blocks, the normal force is equal to the weight of the block i.e., \(mg\). The friction force on the 8 kg block is: \(f_1 = (0.5) (8) (10) = 40 N\), and the friction force on the 4 kg block is: \(f_2 = (0.5) (4) (10) = 20 N\).
02

Analyze forces acting on the 8 kg block

According to Newton's second law, the net force acting on an object is equal to the product of its mass and acceleration, i.e., \(F_{net} = ma\), where \(F_{net}\) is the net force, \(m\) is the mass of the object, and \(a\) is the acceleration. For the 8 kg block, the net force is the difference between the tension in the string (T) and the friction force acting on it. We have: \(F_{net} = T - f_1\).
03

Analyze forces acting on the 4 kg block

For the 4 kg block, the net force is the difference between the force applied (F) and the friction force acting on it, and also the tension acting in the opposite direction. The tension T is in the negative direction with respect to the net force F acting on the 4 kg block, so we have: \(F_{net} = -T + F - f_2\).
04

Apply the condition of constant tension

For the tension in the string to remain constant throughout the motion, the acceleration of both blocks must be equal. Therefore, by equating their net forces we have: \[(T - f_1) = -(-T + F - f_2)\]
05

Solve for F

Now, we will solve this equation to find F: \begin{align*} T - f_1 &= -(-T + F - f_2)\\ 2T - (f_1 + f_2) &= F\\ \end{align*} We know that T is also the force required to overcome the friction acting on the 8 kg block, therefore: \[T = f_1\] Substituting this value of T into the equation, we get: \begin{align*} 2f_1 - (f_1 + f_2) &= F\\ F &= 40 N\\ \end{align*} So, the force required is 40 N. The correct answer is (A) \(40 \mathrm{~N}\).

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Most popular questions from this chapter

A player caught a cricket ball of mass \(150 \mathrm{~g}\) moving at the rate of \(20 \mathrm{~ms}^{-1}\). If the catching process be completed in $0.1 \mathrm{~s}$ the force of the blow exerted by the ball on the hands of player is (A) \(0.3 \mathrm{~N}\) (B) \(30 \mathrm{~N}\) (C) \(300 \mathrm{~N}\) (D) \(3000 \mathrm{~N}\)

Match the column \begin{tabular}{l|l} Column - I & Column - II \end{tabular} (a) Body line on a (p) is a self adjusting horizontal surface \(\quad\) force (b) Static friction (q) is a maximum value of static friction (c) Limiting friction (r) is then limiting friction (d) Dynamic friction (s) force of friction \(=0\) (A) $\mathrm{a}-\mathrm{s}, \mathrm{b}-\mathrm{p}, \mathrm{c}-\mathrm{q}, \mathrm{d}-\mathrm{r}$ (B) $\mathrm{a}-\mathrm{p}, \mathrm{b}-\mathrm{q}, \mathrm{c}-\mathrm{r}, \mathrm{d}-\mathrm{s}$ (C) $\mathrm{a}-\mathrm{s}, \mathrm{b}-\mathrm{r}, \mathrm{c}-\mathrm{q}, \mathrm{d}-\mathrm{p}$ (D) $\mathrm{a}-\mathrm{r}, \mathrm{b}-\mathrm{q}, \mathrm{c}-\mathrm{p}, \mathrm{d}-\mathrm{s}$

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The motion of a particle of a mass \(m\) is describe by \(\mathrm{y}=\mathrm{ut}+(1 / 2) \mathrm{gt}^{2}\). Find the force acting on the particle. (A) \(\mathrm{F}=\mathrm{ma}\) (B) \(\mathrm{F}=\mathrm{mg} \quad\) (C) \(\mathrm{F}=0\) (D) None of these

which of the following statement is correct? (A) A body has a constant velocity but a varying speed. (B) A body has a constant speed but a varying value of acceleration. (C) A body has a constant speed and zero acceleration. (D) A body has a constant speed but velocity is zero.
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