physics question

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

The component of gravitational acceleration down the incline is given by a = g x sinθ. We don't know the incline angle so we can't calculate a using that equation.

However, we can calculate a using this equation, V_f² = V₀² + 2 x a x L where V_f is the final velocity (3.8 m/s), V₀ is the initial velocity (zero), L is the length travelled (6.8 m). Substituting the known values into the equation and solving for a, we get a = 1.06 m/s².

Now we can use the same equation to calculate the speed when the block is 3.6 m from the top of the slide/

In this case, L is 3.6 m, a is 1.06 m/s². The only unknown is V_f.

V_f² = 0 + 2 x 1.06 x 3.6. V_f = 2.76 m/s.

Check math.