A 0.075-kg arrow is fired horizontally. The bowstring exerts a force on the arrow over a distance of 0.90 m. the bow at 40 m/s. What average force does the bow apply to arrow?

The magnitude of average force applied by the bow to the arrow is 66.67 N.

Given data:

The mass of arrow is, m = 0.075 kg.

The distance covered by the arrow is,  s = 0.90 m.

The speed of bow is, v = 40 m/s.

In this problem, the work - energy theorem can be applied. As per the work -energy theorem, the work done by the applied force is equal to the kinetic energy change.

Here, F is the magnitude of average force.

Solving as,

Thus, we can conclude that the magnitude of average force applied by the bow to the arrow is 66.67 N.

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Use work-energy to solve this

The work done by on the bowstring will be converted into kinetic energy of the arrow

so W = F*d = K = 1/2m*v^2

o v = sqrt(2*F*d/m) = sqrt(2*65N*0.90m/0.075kg) = 39.5 m/s

Explanation:

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 .