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Chapter 4: Problem 9

Two blocks of equal mass are connected by a massless horizontal rope and resting on a frictionless table. When one of the blocks is pulled away by a horizontal external force \(\vec{F}\) what is the ratio of the net forces acting on the blocks? a) 1: 1 c) 1: 2 b) 1: 1.41 d) none of the above

Short Answer

Expert verified
Answer: d) none of the above. The ratio of the net forces acting on the two blocks is F - T, where F is the external force and T is the tension force in the rope.

Step by step solution

01

Analyze the forces acting on Block 1

Let's focus on the block that has the external force \(\vec{F}\) applied to it (let's call it Block 1). The forces acting on Block 1 are the external force \(\vec{F}\) and the tension force \(\vec{T}\) due to the rope. Since both forces are in the horizontal direction and opposing each other, we can write Newton's second law for Block 1 as: \(F - T = m_1a_1\) (1)
02

Analyze the forces acting on Block 2

Now let's focus on the other block (let's call it Block 2). The only force acting on it is the tension force \(\vec{T}\) due to the rope. We can write Newton's second law for Block 2 as: \(T = m_2a_2\) (2)
03

Express both accelerations in terms of tension and masses

From equation (1), we can express \(a_1\) in terms of tension and the mass of Block 1: \(a_1 = \frac{F - T}{m_1}\) (3) From equation (2), we can express \(a_2\) in terms of tension and the mass of Block 2: \(a_2 = \frac{T}{m_2}\) (4)
04

Find the ratio of accelerations

Divide equation (4) by equation (3) to find the ratio of \(a_2\) to \(a_1\): \(\frac{a_2}{a_1} = \frac{\frac{T}{m_2}}{\frac{F - T}{m_1}}\) (5) Since both blocks have equal masses (as stated in the exercise), \(m_1 = m_2\), and equation (5) simplifies to: \(\frac{a_2}{a_1} = \frac{T}{F - T}\) (6)
05

Find the ratio of net forces

To find the ratio of the net forces acting on the two blocks, we need to relate their accelerations to the net forces. From Newton's second law, \(F_{net} = ma\), where \(F_{net}\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. Therefore, we have: \(\frac{F_{net1}}{F_{net2}} = \frac{m_1a_1}{m_2a_2}\) (7) Using equation (6) and the fact that both blocks have equal masses, we find: \(\frac{F_{net1}}{F_{net2}} = \frac{m_1a_1}{m_2\frac{T}{F - T}} = \frac{1}{\frac{1}{F - T}} = F - T\) (8)
06

Determine the ratio from the options given

From equation (8), we conclude that the ratio of the net forces acting on the two blocks is \(F - T\). Since \(T\) is always less than \(F\), the ratio cannot be equal to any of the options given (a, b, or c). Therefore, the answer is: d) none of the above

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Most popular questions from this chapter

-4.44 A mass \(m_{1}=20.0 \mathrm{~kg}\) on a frictionless ramp is attached to a light string. The string passes over a frictionless pulley and is attached to a hanging mass \(m_{2}\). The ramp is at an angle of \(\theta=30.0^{\circ}\) above the horizontal. \(m_{1}\) moves up the ramp uniformly (at constant speed). Find the value of \(m_{2}\)

What coefficient of friction is required to stop a hockey puck sliding at \(12.5 \mathrm{~m} / \mathrm{s}\) initially over a distance of \(60.5 \mathrm{~m} ?\)

4.71 A block of mass \(20.0 \mathrm{~kg}\) supported by a vertical massless cable is initially at rest. The block is then pulled upward with a constant acceleration of \(2.32 \mathrm{~m} / \mathrm{s}^{2}\). a) What is the tension in the cable? b) What is the net force acting on the mass? c) What is the speed of the block after it has traveled \(2.00 \mathrm{m?}\)

A \(2.00-\mathrm{kg}\) block is on a plane inclined at \(20.0^{\circ}\) with respect to the horizontal. The coefficient of static friction between the block and the plane is \(0.60 .\) a) How many forces are acting on the block? b) What is the normal force? c) Is this block moving? Explain.

4.57 Your refrigerator has a mass of \(112.2 \mathrm{~kg}\), including the food in it. It is standing in the middle of your kitchen, and you need to move it. The coefficients of static and kinetic friction between the fridge and the tile floor are 0.460 and 0.370 , respectively. What is the magnitude of the force of friction acting on the fridge, if you push against it horizontally with a force of each magnitude? a) \(300 \mathrm{~N}\) b) \(500 \mathrm{~N}\) c) \(700 \mathrm{~N}\)
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