Ask Question
3 March, 15:41

3.3 kg block is on a perfectly smooth ramp that makes an angle of 52° with the horizontal. (a) What is the block's acceleration (in m/s2) down the ramp? (Enter the magnitude.) m/s2 What is the force (in N) of the ramp on the block? (Enter the magnitude.) N (b) What force (in N) applied upward along and parallel to the ramp would allow the block to move with constant velocity?

+3
Answers (1)
  1. 3 March, 17:05
    0
    a) a = 7.72 m / s², N = 19.9 N and b) F = 25.5 N

    Explanation:

    To solve this problem we will use Newton's second law, let's set a reference system with an axis parallel to the plane and gold perpendicular axis. Let's break down the weight (W)

    sin52 = Wx / W

    cos52 = Wy / W

    Wx = W sin52

    Wy = w cos 52

    Let's write them equations

    X axis

    Wx = ma

    Y Axis

    N-Wy = 0

    N = Wy

    a) Let's calculate the acceleration

    a = W sin52 / m = mg sin 52 / m

    a = g sin 52

    a = 9.8 sin52

    a = 7.72 m / s²

    The force of the ramp is normal

    N = Wy = mg cos 52

    N = 3.3 9.8 cos 52

    N = 19.9 N

    b) For the block to move at constant speed the sum of force on the axis must be zero,

    F - Wx = 0

    F = Wx

    F = mg sin52

    F = 3.3 9.8 sin 52

    F = 25.5 N

    Parallel to the plane and going up
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “3.3 kg block is on a perfectly smooth ramp that makes an angle of 52° with the horizontal. (a) What is the block's acceleration (in m/s2) ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers