23.03.2023 - 03:13

Problem 1. A tetherball leans against the smooth, frictionless post to which it is attached. The string is attached to the ball such that a line along the string passes through the center of the ball.

Question:

Problem 1. A tetherball leans against the smooth, frictionless post to which it is attached. The string is attached to the ball such that a line along the string passes through the center of the ball. The string is 1.30 m long and the ball has a radius of 0.160 m with mass 0.470 kg.

Part A: What is the tension in the rope?

Part B: What is the force the pole exerts on the ball?

Problem 2. An 88 N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward.

Calculate the coefficient of kinetic friction between the box and floor. Express your answer using two significant figures.

Answers (1)
  • Vesta
    April 1, 2023 в 11:20

    Part A: To find the tension in the rope, we can use the equation for centripetal force:

    Fc = mv^2/r

    Where Fc is the centripetal force, m is the mass of the ball, v is the velocity of the ball (which in this case is 0 since it is at rest), and r is the radius of the circle the ball would travel if it were released. Since the ball is stationary, we can use the weight of the ball as the force pushing it towards the post, which is equal and opposite to the tension in the rope:

    T = mg = (0.470 kg)(9.81 m/s^2) = 4.61 N

    Therefore, the tension in the rope is 4.61 N.

    Part B: Since the post is smooth and frictionless, the force exerted on the ball by the pole must be perpendicular to the surface of the post. This force is equal and opposite to the tension in the rope, so it is also 4.61 N.

    To find the coefficient of kinetic friction between the box and floor, we can use the equation for acceleration:

    a = Fnet/m

    Where a is the acceleration of the box, Fnet is the net force acting on the box, and m is the mass of the box. We can find the net force by subtracting the force of friction from the push force:

    Fnet = Fpush - Ffriction

    Since the box is slowing down, the direction of the net force is opposite to the direction of motion. We can use the vertical component of the push force to find the normal force acting on the box:

    N = mg - Fdown = (88 N)(9.81 m/s^2) - (25 N) = 857.08 N

    Where Fdown is the force of gravity pulling the box downward. Now we can find the force of friction using the coefficient of kinetic friction, µk:

    Ffriction = µkN

    Substituting back into the equation for net force:

    Fnet = (20 N - µkN) - (88 N)(0.90 m/s^2) = -80.8 N

    Since the acceleration is in the opposite direction of the push force, we can use the absolute value of the acceleration:

    a = -0.90 m/s^2

    Now we can solve for the coefficient of kinetic friction:

    µk = Ffriction/N = (20 N + 80.8 N)/857.08 N = 0.112

    Therefore, the coefficient of kinetic friction between the box and floor is 0.11 (to two significant figures).

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