Question

A small block is released from rest at the top of a frictionless incline. The distance from the top of the incline to the bottom, measured along the incline, is 3.40 m. The vertical distance from the top of the incline to the bottom is 1.09 m. If g

Answer 1

The subject of the question is Physics, where principles of kinematics and Newton's laws of motion are applied to determine the acceleration and motion of a block sliding down an incline.

Step-by-step explanation:

When considering a block sliding down a frictionless incline and the effects of gravity on the motion of bodies, we're dealing with a physics problem. The motion of the block can be analyzed using the principles of kinematics and Newton's laws of motion. For instance, assuming a frictionless surface, if a block is released from rest, it will accelerate down the incline due to gravity. The component of gravitational force acting along the incline is what propels the block forward. This scenario is common in physics problems which invoke the use of equations of motion to find various quantities like acceleration, distances, or forces acting upon the objects.

Answer 2

Complete Question

A small block is released from rest at the top of a frictionless incline. The distance from the top of the incline to the bottom, measured along the incline, is 3.40 m. The vertical distance from the top of the incline to the bottom is 1.09 m. If
`g=9.8 \ m/s^2` , what is the acceleration of the block as it slides down the incline plane

Answer:

The acceleration is
`a = 3.142 m/s^2`

Step-by-step explanation:

From the question we are told that

The distance from top to bottom of the inclined plane measured along the incline is
`d = 3.40 \ m`

The distance from top to bottom of the inclined plane measured along the vertical axis is

`D = 1.90 \ m`

According to the SOHCAHTOA rule

`sin \theta = (D)/(d)`

=>
`\theta = sin ^(-1) [(D)/(d) ]`

substituting values

=>
`\theta = sin ^(-1) [(1.09)/(3.40) ]`

`\theta = 18.699^o`T

The acceleration of a block on a frictionless inclined plane is mathematically represented as

`a = gsin \theta`

substituting values

`a = 9.8 * sin(18.699)`

`a = 3.142 m/s^2`