/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 100 A block \(P\) of mass \(4 \mathr... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 100

A block \(P\) of mass \(4 \mathrm{~kg}\) is placed on horizontal rough surface with co-efficient of friction \(\mu=0.6\). And two blocks \(R\) and \(Q\) of masses \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) connected with the help of massless strings \(A\) and \(B\), respectively, passing over frictionless pulleys as shown, then \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) Acceleration of block \(P\) is zero. (B) Tension in string \(A\) is \(20 \mathrm{~N}\). (C) Tension in string \(B\) is \(40 \mathrm{~N}\). (D) Contact force on block \(P\) is \(20 \sqrt{5} \mathrm{~N}\).

Short Answer

Expert verified
The short answer is: The acceleration of block P is \(9\,\frac{\text{m}}{\text{s}^2}\), tension in string A is \(20\,\text{N}\), and tension in string B is \(40\,\text{N}\). Therefore, statements B and C are true, while statements A and D are false.

Step by step solution

01

Analyze the forces acting on each block

First of all, we are going to analyze the forces acting on each block. For block P, there is a gravitational force W = mg acting downward and a normal force N acting upward, and a friction force f = 渭N acting horizontally to the left. For blocks Q and R, there is a gravitational force acting downward and the tension in the strings (TA and TB) acting upward.
02

Calculate the maximum friction force for block P

Since block P is on a horizontal surface, the normal force N is equal to the gravitational force W for this block: \(N = W = mg = (4\,\text{kg})(10\,\frac{\text{m}}{\text{s}^2}) = 40\,\text{N}\) Now we'll calculate the maximum friction force that can act on block P: \(f_\text{max} = \mu N = (0.6)(40\,\text{N}) = 24\,\text{N}\)
03

Calculate the net force acting on block Q

There are two forces acting on block Q: the gravitational force and the tension in string A. So we can write the following equation: \(T_A = m_Qg \Rightarrow T_A = (2\,\text{kg})(10\,\frac{\text{m}}{\text{s}^2}) = 20\,\text{N}\)
04

Calculate the net force acting on block R

There are also two forces acting on block R: the gravitational force and the tension in string B. So we can write the following equation: \(T_B = m_Rg \Rightarrow T_B = (4\,\text{kg})(10\,\frac{\text{m}}{\text{s}^2}) = 40\,\text{N}\)
05

Calculate the net force acting on block P and its subsequent acceleration

For block P, there is a horizontal net force acting to the right with a magnitude equal to the tensions in the strings: \(F_\text{net} = T_A + T_B = 20\,\text{N} + 40\,\text{N} = 60\,\text{N}\) Since the net force on block P (60 N) is greater than the maximum friction force (24 N), the block will accelerate and not remain stationary. Let's now find the acceleration or block P: \(a_P = \frac{F_\text{net} - F_\text{friction}}{m} = \frac{60\,\text{N} - 24\,\text{N}}{4\,\text{kg}} = \frac{36\,\text{N}}{4\,\text{kg}} = 9\,\frac{\text{m}}{\text{s}^2}\)
06

Check the given statements

Now, let's verify if the given statements are true or false based on our calculations: (A) Acceleration of block P is zero. [False] - We found that the acceleration of block P is 9 m/s虏. (B) Tension in string A is 20 N. [True] - We calculated the tension in string A to be 20 N. (C) Tension in string B is 40 N. [True] - We calculated the tension in string B to be 40 N. (D) Contact force on block P is \(20\sqrt{5}\,\text{N}\). [False] - The contact force is the normal force N, which is equal to 40 N, not \(20\sqrt{5}\,\text{N}\). In conclusion, statements B and C are true, while statements A and D are false.

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Most popular questions from this chapter

The upper half of an incline plane with inclination \(\phi\) is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the co-efficient of friction for the lower half is given by (A) \(2 \tan \phi\) (B) \(\tan \phi\) (C) \(2 \sin \phi\) (D) \(2 \cos \phi\)

A block of mass \(m\) is placed on the top of another block of mass \(M\) as shown in the Fig. \(3.81\). The co-efficient of friction between them is \(\mu .\) The maximum acceleration with which the block \(M\) may move so that \(m\) also moves along with it is (A) \(\mu g\) (B) \(g / \mu\) (C) \(\mu^{2} / g\) (D) \(g / \mu^{2}\)

A girl of mass \(50 \mathrm{~kg}\) stands on a measuring scale in a lift. At an instant, it is detected that the reading reduces to \(40 \mathrm{~kg}\) for a while and then returns to original value. It can be said that (A) The lift was in constant motion upwards (B) The lift was in constant motion downwards (C) The lift was suddenly started in downward motion (D) The lift was suddenly started in upward motion

A body of mass \(m\) is kept stationary on a rough inclined plane of inclination \(\theta .\) The magnitude of force acting on the body by the inclined plane is (A) \(m g\) (B) \(m g \sin \theta\) (C) \(m g \cos \theta\) (D) \(m g \sqrt{1+\cos ^{2} \theta}\)

A constant force acting on a body of mass \(3.0 \mathrm{~kg}\) changes its speed from \(2.0 \mathrm{~ms}^{-1}\) to \(3.5 \mathrm{~ms}^{-1}\) in \(25 \mathrm{~s}\). The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force? (A) \(0.18 \mathrm{~N}\) (B) \(0.36 \mathrm{~N}\) (C) \(0.9 \mathrm{~N}\) (D) None of these
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