The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at position x=x1x=x1

Question

The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at position x=x1x=x1. Find FfFfF_f, the magnitude of the average frictional force that acts on the box. (Since you don’t know the coefficient of friction, don’t include it in your answer.) Express the frictional force in terms of mmm, v0v0v_0, and x1x1x_1.

in progress 0
niczorrrr 3 years 2021-08-05T19:26:30+00:00 1 Answers 51 views 0

Answers ( )

  1. trucchi
    0
    2021-08-05T19:27:39+00:00
    Reply

    Answer:

    fr = ½ m v₀²/x

    Explanation:

    This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

    The best way to solve this exercise is to use the energy work theorem

                W = ΔK

    Where work is defined as the product of force by distance

               W = fr x cos 180

    The angle is because the friction force opposes the movement

              Δk =K_{f} –K₀

              ΔK = 0 – ½ m v₀²

    We substitute

             – fr x = – ½ m v₀²      

               fr = ½ m v₀²/x

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )