Aman decides to push a 2.0 kg box from the bottom to the top of a 30-degree incline the coeficient of kinetic friction is 0.2 waht is the amount of force required to keep the box moving up the incline at a constant

Explanation:

m = 2 Kg

α = 30º

μ = 0.2

F = ?

We can apply

∑ Fx' = 0   (as v is constant, a = 0)

then

F - Ff - Wx' = 0    ⇒   F = Ff + Wx'

We get the force of friction as follows

Ff = μ * N = μ * Wy' = μ * (W * Cosα) = μ * m * g * Cosα  

⇒  Ff = 0.2 * 2 Kg * 9.81 m/s² * Cos30º = 3.3983 N

then we get the x'-component of W

Wx' = W * Sin α = m * g * Sin α

⇒  Wx' = 2 Kg * 9.81 m/s² * Sin30º = 9.81 N

finally we obtain

F = Ff + Wx' = 3.3983 N + 9.81 N = 13.2083 N

we have been given a quadratic function 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 .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .