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Physics Asked by a SnapAndSolve user · about 2 hours ago

1 The coefficient of friction for the surface, for both objects is 0,20 and the system accelerates to the...

5.1 The coefficient of friction for the surface, for both objects is 0,20 and the system accelerates to the right at 2,08 m·s⁻².
5.1.1 State Newton’s Second Law of motion in words. (2)
5.1.2 Draw a fully labelled free-body diagram of ALL the forces acting on block A. (5)
5.1.3 Show that the frictional force experienced by block A while accelerating is 17,4 N. (4)
5.1.4 Calculate the tension force in the string between block A and B. (5)

Answer

  1. Newton’s Second Law of Motion: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The direction of the acceleration is in the direction of the net force acting on the object.

  2. Free-body diagram for block A:
    Forces acting on block A include:
    - The applied force of 120 N at an angle of 30° to the horizontal.
    - The gravitational force (weight) acting downward.
    - The normal force acting upward.
    - The frictional force acting opposite to the direction of motion.

  3. Frictional force on block A:
    The normal force (N) on block A is equal to the gravitational force minus the vertical component of the applied force.
    N = mAg - 120 sin(30°)
    N = 15 kg × 9.8 m/s² - 120 × 0.5
    N = 147 N - 60 N = 87 N
    The frictional force (f) is given by f = μN.
    f = 0.20 × 87 N = 17.4 N

  4. Tension force in the string:
    Let T be the tension in the string. The net force on block A is given by:
    Fnet = mAa = 120 cos(30°) - T - f
    15 kg × 2.08 m/s² = 120 × √3/2 - T - 17.4 N
    31.2 N = 103.92 N - T - 17.4 N
    T = 103.92 N - 17.4 N - 31.2 N
    T = 55.32 N

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