Question

A small block with mass 0.320 kg is sliding down a frictionless ramp that is inclined at an angle of 53.1° above the horizontal. assume g = 9.80 m/s² . as the object slides down the incline, what is the magnitude of the component of the gravity force parallel to the surface of the incline?

Answer 1

The component of the gravitational force parallel to the incline is found using the formula F_parallel = m × g × sin(θ). Substituting the given values, the magnitude of this component is approximately 2.51 N.

Step-by-step explanation:

To calculate the magnitude of the component of the gravitational force that is parallel to the incline, we use:

Fparallel = m × g × sin(θ)

Where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the incline above the horizontal plane. For the given problem:

• 
  
• 
  
• 
  

Plugging in the values, we find:

Fparallel = 0.320 kg × 9.80 m/s² × sin(53.1°)

First, calculate the sine of the angle:

sin(53.1°) ≈ 0.7986

Now, multiply the mass, acceleration due to gravity, and the sine of the angle:

Fparallel = 0.320 kg × 9.80 m/s² × 0.7986 ≈ 2.51 N

The magnitude of the component of the gravitational force parallel to the surface of the incline is approximately 2.51 N.