physics question 13

Question

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled 6.80 mm to the bottom of the incline is 3.80 m/sm/s.

What is the speed of the block when it is 3.60 mm from the top of the incline?

Answer 1

Solution: I am assuming that the velocity at the bottom of the incline is 3.8 mm/s

For a moving body with constant acceleration a = dv / dt = cte

The block is released from rest meaning that at t = 0, v =0, and at a distance of 6.80 mm from the top of the incline the velocity is 3.8 mm/s. Using the equation v_f² - v_i² = 2ax determine the acceleration

substituting: v_i = 0, v_f = 3.8 mm/s, x = 6.8 mm

a = (v_f² - v_i²) / 2x = [(3.8 mm/s)² - 0] / 2( 6.8mm) = 1.062 mm/s²

Since the acceleration is constant at 1.062 mm/s², the velocity of the block when it is 3.60 mm from the top of the incline can be found from the same equation, here v_i =0 and we have to find v_f

and v_f² - v_i² = 2ax, v_f² - 0 = 2( 1.062 mm/s²)( 3.6 mm), v_f² = 7.65 mm²/ s² , v_f = 2.77 mm/s