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

magnitude and direction

1/0.14nm = 10^10/1.4 a lot