Physics, 19.12.2019 07:31 oliwia0765

If a box is not accelerating on a slope of 20 degrees, what is the net force acting on the box?

if it is accelerating, what is the net force?

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Answer from: natperal

The net force is zero, if there is not external forces. The net force if the body is accelerating will be the product of the mass by the acceleration.

Explanation:

If the box does not accelerate, there are two possible options, either the box moves at constant velocity or does not receive any external force to alter its balance.

Newton's Second Law, also known as the Fundamental Law of dynamics, determines a proportional relationship between Force and variation of the amount of motion or linear moment of a body. In other words, The Force is directly proportional to the mass and acceleration of a body.

$F=m*a\\where:\\m= mass [kg]\\a=acceleration[m/s^2]$

$0=m*a\\a=0[m/s^2]$

If there are forces acting on the body and these forces are not in balance the body will move with a resulting force equal to the product of mass by acceleration.

$F_{net} =m*a$

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

Answer from: Quest

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If a box is not accelerating on a slope of 20 degrees, what is the net force acting on the box?

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