By how many percent does a 40.0 degree inclined plane reduce the normal force of a 380 N object?
cos 40 = .766
so
100 - 76.6
that's not the answer
To find the reduction in the normal force on an object on a inclined plane, we need to calculate the component of the weight force that acts perpendicular to the plane.
Let's start by finding the weight force acting on the object. The weight force of an object can be calculated using the formula:
Weight = mass × acceleration due to gravity
Since we are given the value of the normal force, we can use the equation for the normal force to find the mass of the object. The equation for the normal force is:
Normal force = mass × gravity
We can rearrange the equation to solve for mass:
mass = Normal force / gravity
Substituting the given values:
mass = 380 N / 9.8 m/s^2 = 38.78 kg
Now, let's find the component of the weight force that acts perpendicular to the plane. This component is given by:
Component = Weight × sin(angle)
Substituting the values:
Component = (mass × gravity) × sin(angle)
Component = (38.78 kg × 9.8 m/s^2) × sin(40.0 degrees)
Component ≈ 235.3 N
The reduction in the normal force can be calculated by subtracting the component from the weight force:
Reduction = Weight - Component
Reduction = 380 N - 235.3 N
Reduction ≈ 144.7 N
To find the reduction in terms of a percentage, we need to calculate the percentage of the reduction compared to the original normal force. This can be found using the formula:
Percentage reduction = (Reduction / Normal force) × 100%
Substituting the values:
Percentage reduction = (144.7 N / 380 N) × 100%
Percentage reduction ≈ 38.08%
Therefore, a 40.0 degree inclined plane reduces the normal force of a 380 N object by approximately 38.08%.