/*! 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 152 A block of mass \(M\) is pulled ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 152

A block of mass \(M\) is pulled along a horizontal frictionless surface by a rope of mass \(m\). If a force \(P\) is applied at the free end of the rope, the force exerted by the rope on the block is (A) \(\frac{P m}{M+m}\) (B) \(\frac{P m}{M-m}\) (C) \(P\) (D) \(\frac{P M}{M+m}\)

Short Answer

Expert verified
\(F = \frac{P M}{M+m}\)

Step by step solution

01

Analyze forces on the rope and block separately

As we are dealing with two separate objects (the rope and the block), we need to analyze the forces acting on each of them separately. We will first consider the rope and then the block.
02

Apply Newton's second law for the rope

For the rope, the applied force P acts on it, and a force F (the force exerted by the rope on the block) acts in the opposite direction. According to Newton's second law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration. So, for the rope we have: \(P - F = m a\)
03

Apply Newton's second law for the block

For the block, the only horizontal force acting on it is the force exerted by the rope (F). Therefore, Newton's second law for the block is: \(F = M a\)
04

Solve for the force exerted by the rope on the block (F)

Now, we have two equations with the force F and the acceleration a: 1. \(P - F = m a\) 2. \(F = M a\) We can solve for F by eliminating the variable a. To do this, first, solve equation 2 for a, getting: \(a = \frac{F}{M}\) Now substitute this expression for a into equation 1: \(P - F = m (\frac{F}{M})\) Now solve for F. First, multiply both sides by M, yielding: \(P M - F M = m F\) Next, move the term with F to one side of the equation: \(P M = F(M + m)\) Finally, divide both sides by (M + m) to get F: \(F = \frac{P M}{M+m}\) So, the force exerted by the rope on the block is: \(F = \frac{P M}{M+m}\) Therefore, the correct answer is: (D) \(\frac{P M}{M+m}\)

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Most popular questions from this chapter

One end of a string of length \(l\) is connected to a particle of mass \(m\) and the other to a small peg on a smooth horizontal table. If the particle moves in a circular motion with speed \(v\), the net force on the particle (directed towards the centre) is: (A) \(T\) (B) \(T-\frac{m v^{2}}{l}\) (C) \(T+\frac{m v^{2}}{l}\) (D) 0

A \(2 \mathrm{~kg}\) stone at the end of a string \(1 \mathrm{~m}\) long is whirled in a vertical circle at a constant speed. The speed of the stone is \(4 \mathrm{~m} / \mathrm{s}\). The tension in the string will be \(52 \mathrm{~N}\), when the stone is (A) at the top of the circle (B) at the bottom of the circle (C) halfway down (D) none of the above

The magnitude and direction of the net force acting on a stone of mass \(0.1 \mathrm{~kg}\) is given below, choose the incorrect statement. (A) Just after it is dropped from the window of a stationary train, \(F=1.0 \mathrm{~N}\) (Vertically down ward) (B) Just after it is dropped from the window of a train running at a constant velocity of \(36 \mathrm{~km} / \mathrm{h}\), \(F=1.0 \mathrm{~N}\) (Vertically downward). (C) Just after it is dropped from the window of a train accelerating with \(1 \mathrm{~ms}^{-2}, F=1.0 \mathrm{~N}\) (Vertically downward). (D) Just after it is dropped from the window of a train accelerating with \(1 \mathrm{~ms}^{-2}, F=2.0 \mathrm{~N}\) (Vertically downward).

Two blocks of masses \(5 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) are placed on a smooth horizontal surface. A horizontal force \(F=16 \mathrm{~N}\) is applied on \(5 \mathrm{~kg}\) as shown. Find normal force between the blocks.

Two bodies of masses \(10 \mathrm{~kg}\) and \(20 \mathrm{~kg}\), respectively, kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force \(F=600 \mathrm{~N}\) is applied to body of mass \(10 \mathrm{~kg}\). What is the tension in the string in each case? (A) \(200 \mathrm{~N}\) (B) \(100 \mathrm{~N}\) (C) \(400 \mathrm{~N}\) (D) \(600 \mathrm{~N}\)
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