Question

a 2 kg block, starting from rest, slides 20 m down a frictionless inclined plane from x to y, dropping a vertical distance of 10 m as shown above. question the magnitude of the net force on the block while it is sliding is most nearly responses 0.1 n 0.1 n 0.4 n 0.4 n 2.5 n 2.5 n 5.0 n 5.0 n 10.0 n

Answer 1

The net force on a 2 kg block sliding down a frictionless inclined plane, which allows for a 10 m vertical drop, is about 10.0 N. This is calculated by finding the component of the gravitational force acting down the slope.

Step-by-step explanation:

The magnitude of the net force on a block sliding down a frictionless inclined plane can be found using Newton's second law of motion and the component of the gravitational force acting down the slope. Since there is no friction, the only force acting along the incline is the component of the weight of the block parallel to the surface of the incline (mg sinθ), where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of inclination.

In this scenario, with a mass (m) of 2 kg, a gravitational acceleration (g) of 9.8 m/s², and a vertical drop of 10 m which corresponds to an angle of inclination (θ), we can calculate the angle using trigonometric relationships. Specifically, sinθ = opposite/hypotenuse, which gives us sinθ = 10 m / 20 m =0.5. This means θ = 30°.

Applying the formula mg sinθ to find the force, we get F = 2 kg * 9.8 m/s² * sin(30°), or approximately 9.8 N. Thus, the magnitude of the net force on the block while it is sliding is most nearly 10.0 N.

Answer 2

the magnitude of the net force on the block while it is sliding is most nearly responses is
`\( 10.0 \, \text{N} \).`

To find the magnitude of the net force on the block while it is sliding down a frictionless inclined plane, we can use the principles of physics, particularly Newton's second law and the concept of gravitational force on an inclined plane. Here's how to approach this problem step by step:

Step 1: Understand the Scenario

- You have a 2 kg block sliding down a frictionless inclined plane.

- The vertical drop is 10 m.

Step 2: Calculate the Gravitational Force Component Along the Incline

- The gravitational force acting on the block is
`\( F_g = m * g \),` where m is the mass and g is the acceleration due to gravity (approximately
`\( 9.8 \, \text{m/s}^2 \)).`

- On an inclined plane, only the component of gravitational force parallel to the plane affects the movement of the block. This component can be found using the sine of the angle of inclination
`(\( \theta \)).`

- The angle
`\( \theta \)` can be determined using trigonometry, since we know the vertical drop and the hypotenuse (the length of the incline):

`\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\text{vertical drop}}{\text{length of incline}} \]`

The force down the incline (parallel component) is given by:

`\[ F_(\parallel) = m * g * \sin \theta \]`

Step 3: Calculate the Force

First, find
`\( \sin \theta \): \[ \sin \theta = \frac{10 \, \text{m}}{20 \, \text{m}} = 0.5 \]`

Calculate the gravitational force component along the incline:

`\[ F_(\parallel) = 2 \, \text{kg} * 9.8 \, \text{m/s}^2 * 0.5 \]`

Step 4: Evaluate the Expression

Let's calculate the final value of the force
`\( F_(\parallel) \).`

The magnitude of the net force on the block while it is sliding down the inclined plane is approximately
`\( 9.8 \, \text{N} \).` Therefore, the closest answer among the given options is
`\( 10.0 \, \text{N} \).`