Ablock of mass 3m is released from rest at a height r above a horizontal surface. theacceleration due to gravity is g. the block slides along the inside of a frictionless circular hoop of radius r. what is the magnitude of the normal force exerted on the block by the hoop when the block has reached the bottom?

N = 3 mg

Explanation:

For this case we will use Newton's second law where acceleration is centripetal.

    F = m a

    a = v² / r

Let's write the expression for the lowest part of the hoop

    N- W = m a

    N = mg + m V2 / r

To find the speed let's use energy conservation

highest point

     Em₀ = U = m g y

lowest point

      = K = ½ m v²

    Em₀ =  

    mg y = ½ m v²

    v = √ 2gy

We substitute in the equation of the normal

    N = m (g 2g y/r )

Let's analyze the initial height it gives is that the body leaves a height R

    N = m (g + 2 g R / R)

     N = 3 mg

neither accurate nor precise.

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.