Question

PLEASE HELP ASP!!!!! WILL GIVE POINTS AND BRAINLIEST!!!!!!!!
I only need answer for the second and third question. The answer for the first question is 3.8291.

2. What is the block’s speed when, starting from rest, it has traveled a distance of 2.41 m along the incline?

3. What is the magnitude of the perpendicular force that the block exerts on the surface of the plane at a distance of 2.41 m down the incline?

Answer 1

Answer:

Explanation:

Agree with your answer for the first part, though I would round to 3.83 m/s²

2) v² = u² + 2as

   v² = 0² + 2(3.8291)(2.41)

   v² = 18.456575...

    v = 4.296111...

    v = 4.30 m/s

3) It does not matter where on the plane it is calculated, the normal force will be constant at Fn = mgcosθ

  F = 2.47(9.8)cos23

  F = 22.28174...

  F = 22.3 N