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Chapter 4: Problem 48

A rectangular block of width \(w=116.5 \mathrm{~cm},\) depth \(d=164.8 \mathrm{~cm}\) and height \(h=105.1 \mathrm{~cm}\) is cut diagonally from one upper corner to the opposing lower corners so that a triangular surface is generated, as shown in the figure. A paperweight of mass \(m=16.93 \mathrm{~kg}\) is sliding down the incline without friction. What is the magnitude of the acceleration that the paperweight experiences?

Short Answer

Expert verified
Answer: The magnitude of the acceleration that the paperweight experiences while sliding down the incline without friction is approximately \(4.50\mathrm{~m/s^2}\).

Step by step solution

01

Find the length of the diagonal

To find the length of the diagonal, we will use the Pythagorean theorem in 3D space which can be stated as: \(l^2 = w^2 + d^2 + h^2\). Given the dimensions of the rectangular block, we can plug the values and find the diagonal length. \(l^2 = 116.5^2 + 164.8^2 + 105.1^2\) Calculating the diagonal length: \(l = \sqrt{116.5^2 + 164.8^2 + 105.1^2} \approx 230.37 \mathrm{~cm}\)
02

Find the angle of the incline

Next, we need to find the angle at which the incline is tilted. We can use the height \(h\) and the diagonal length \(l\) to find the angle using trigonometry. Let \(\theta\) be the angle between the diagonal \(l\) and the base of the rectangular block. Using the sine function, we can write: \(\sin{\theta} = \frac{h}{l}\) Now, we can find the angle \(\theta\): \(\theta = \arcsin{\frac{105.1}{230.37}} \approx 27.51^\circ\)
03

Find the acceleration of the paperweight along the incline

Finally, we can find the acceleration of the paperweight sliding down the incline using the gravitational force and the angle of the incline. We know that the force acting along the incline is: \(F = mg\sin{\theta}\) Now, we can use Newton's second law \(F = ma\) to find the acceleration: \(16.93\mathrm{~kg} \cdot a = 16.93\mathrm{~kg} \cdot 9.81\mathrm{~m/s^2} \cdot \sin{27.51^\circ}\) Divide both sides by the mass \(m\): \(a = 9.81\mathrm{~m/s^2} \cdot \sin{27.51^\circ} \approx 4.50\mathrm{~m/s^2}\) The magnitude of the acceleration that the paperweight experiences while sliding down the incline without friction is approximately \(4.50\mathrm{~m/s^2}\).

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Most popular questions from this chapter

In a physics class, a 2.70 - g ping pong ball was suspended from a massless string. The string makes an angle of \(\theta=15.0^{\circ}\) with the vertical when air is blown horizontally at the ball at a speed of \(20.5 \mathrm{~m} / \mathrm{s}\). Assume that the friction force is proportional to the squared speed of the air stream. a) What is the proportionality constant in this experiment? b) What is the tension in the string?

Two blocks of masses \(m_{1}\) and \(m_{2}\) are suspended by a massless string over a frictionless pulley with negligible mass, as in an Atwood machine. The blocks are held motionless and then released. If \(m_{1}=3.50 \mathrm{~kg}\). what value does \(m_{2}\) have to have in order for the system to experience an acceleration \(a=0.400 g\) ? (Hint: There are two solutions to this problem.

Leonardo da Vinci discovered that the magnitude of the friction force is usuzlly simply proportional to the magnitude of the normal force; that is, the friction force does not depend on the width or length of the contact area. Thus, the main reason to use wide tires on a race car is that they a) Iook cool. b) have more apparent contact area. c) cost more. d) can be made of softer materials.

True or false: A physics book on a table will not move at all if and only if the net force is zero.

The Tornado is a carnival ride that consists of vertical cylinder that rotates rapidly aboat its vertical axis. As the Tornado rotates, the riders are presscd against the inside wall of the cylinder by the rotation, and the floon of the cylinder drops away. The force that points upward. preventing the riders from falling downward, is a) friction force c) gravity. b) a normal force. d) a tension force.
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