Question

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

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

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: \(l = \sqrt{116.5^2 + 164.8^2 + 105.1^2} \approx 230.37 \mathrm{~cm}\)

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 \(\theta\): \(\theta = \arcsin{\frac{105.1}{230.37}} \approx 27.51^\circ\)

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 \(4.50\mathrm{~m/s^2}\).

Key concepts

Pythagorean theorem in 3D

When working with objects in three-dimensional space, like our rectangular block, the standard Pythagorean theorem gets an extra dimension. For a 3D object, the theorem helps us find diagonals by considering all three spatial dimensions. In our case, we look at width (\(w\)), depth (\(d\)), and height (\(h\)) of the block.

Here's the formula you would use: \[ l^2 = w^2 + d^2 + h^2 \] This tells us that the square of the diagonal length \(l\) is equal to the sum of the squares of its width, depth, and height.

• For width, we have \(116.5 \, \mathrm{cm}\).
• For depth, it is \(164.8 \, \mathrm{cm}\).
• For height, it is \(105.1 \, \mathrm{cm}\).

By calculating \(l^2 = 116.5^2 + 164.8^2 + 105.1^2\) and taking the square root, we find the diagonal \(l\) to be approximately \(230.37 \, \mathrm{cm}\). This diagonal becomes our incline for the paperweight to slide down.

Trigonometry in physics

Trigonometry helps us make sense of angles formed in physical objects and effectively solve physics problems. Here, we use trigonometry to understand the incline, the angle of the diagonal plane with reference to the horizontal base of the block. Let's recall our diagonal length (\(l\)) and height (\(h\)).

The sine function is particularly useful here: \[ \sin{\theta} = \frac{h}{l} \] This equation helps us calculate the angle \(\theta\) of the incline. Trigonometry tells us that this angle is the arc sine of the height divided by the diagonal length.By plugging the values into the formula: \(\theta = \arcsin{\frac{105.1}{230.37}}\), we compute the angle to approximately \(27.51^\circ\). This angle tells us how steep the incline is, which directly affects the subsequent acceleration calculations.

Newton's second law

Newton's second law is a cornerstone in physics, especially when it comes to analyzing motion. The law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

In equation form, that's: \[ F = ma \] In our scenario, the paperweight slides down a frictionless incline. This motion is solely driven by gravity acting parallel to the incline.First, we compute the force down the slope with: \(F = mg \sin{27.51^\circ}\)

• Here, \(m\) is the mass \(16.93 \, \mathrm{kg}\).
• \(g\) is gravitational pull \(9.81 \, \mathrm{m/s^2}\).
• \(\sin{27.51^\circ}\) gives the component of gravitational force along the incline.

By solving for \(a\) using \(F = ma\), and substituting \(F = mg \sin{\theta}\), we divide by \(m\), resulting in: \[ a = 9.81 \, \mathrm{m/s^2} \cdot \sin{27.51^\circ} \approx 4.50 \, \mathrm{m/s^2} \] Thus, the paperweight experiences an acceleration of about \(4.50 \, \mathrm{m/s^2}\) down the incline.