A 9.0-kg box of oranges slides from rest down a frictionless incline from a height of 5.0 m. A constant frictional force, introduced at point A, brings the block to rest at point B, 19 m to the right of point A. What is the speed of the box just before it reaches point A

I'm going to assume that, judging by the fact that the coefficient of friction and angle of the incline seem to be intentionally left out, the point A refers to a point at the bottom of the incline, and the point B is located 19 m away from A on a horizontal surface. If this is not the case, please be sure to correct me!

Let v denote the speed of the box at point A, x the distance the box slides (without friction) down the incline, and θ the angle of the incline.

Recall that

v ² - v₀² = 2 a ∆x

where

v = final speed

v₀ = initial speed

a = acceleration

∆x = displacement

While on the incline, the box starts at rest (v₀ = 0) and slides a distance x from the starting point to point A with acceleration a to attain some speed v.

Use the free body diagram to determine the acceleration. By Newton's second law, the net force on the object acting parallel to the incline

∑ F = m g sin(θ) = m a   →   a = g sin(θ)

where m = 9.0 kg and g = 9.8 m/s².

Write the distance x in terms of θ with some trigonometry:

sin(θ) = (5.0 m) / x   →   x = (5.0 m) / sin(θ)

Now plug a and x into the formula above to solve for v :

v ² = 2 g sin(θ) ((5.0 m) / sin(θ))

v ² = (10.0 m) g

v ≈ 9.9 m/s

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft