A horizontal force of 93.7 N is applied to a 42.5 kg crate on a rough, level surface. If the crate accelerates at 1.03 m/s2, what is the magnitude of the force of kinetic friction (in N) acting on the crate

49.925N

Explanation:

According to newton's second law of motion:

is the sum of force along the x component

m is the mass of the crate

ax is the acceleration

Fk is the magnitude of the force of kinetic friction

Given

Fm = 93.7

m = 42.5kg

a = 1.03m/s²

Substitute into the formula:

Hence the magnitude of the force of kinetic friction (in N) acting on the crate is 49.925N

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 .