The 20 g bullet is traveling at 400 m/s when it becomes embedded in the 2 -kg stationary block. Determine the distance the block will slide

Question

The 20 g bullet is traveling at 400 m/s when it becomes embedded in the 2 -kg stationary block. Determine the distance the block will slide before it stops. The coefficient of kinetic friction between the block and the plane is μk=0.2 .

in progress 0
Thành Công 3 years 2021-08-26T10:42:39+00:00 1 Answers 32 views 0

Answers ( )

  1. sori
    0
    2021-08-26T10:43:55+00:00
    Reply

    Answer:

    The block+bullet system moves 4 m before being stopped by the frictional force.

    Explanation:

    Using the law of conservation of llinear momentum and the work energy theorem, we can obtain this.

    According to Newton’s second law of motion

    Momentum before collision = Momentum after collision

    Momentum before collision = (0.02×400) + 0 (stationary block)

    Momentum before collision = 8 kgm/s

    Momentum after collision = (2+0.02)v

    8 = 2.02v

    v = 3.96 m/s.

    According to the work-energy theorem,

    The kinetic energy of the block+bullet system = work done by Friction to stop the motion of the block+bullet system

    Kinetic energy = (1/2)(2.02)(3.96²) = 15.84 J

    Work done by the frictional force = F × (distance moved by the force)

    F = μmg = 0.2(2.02)(9.8) = 3.96 N

    3.96d = 15.84

    d = (15.84/3.96) = 4 m

Leave an answer

Browse

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