While the box of mass 3.0kg is moving at the constant velocity of 0.10 m/s , force F increases to 18.0 N and acts for 0.20 s. The frictional

Question

While the box of mass 3.0kg is moving at the constant velocity of 0.10 m/s , force F increases to 18.0 N and acts for 0.20 s. The frictional force does not change. Calculate the acceleration produced by the resultant force acting on the box.​

in progress 0
Kiệt Gia 4 years 2021-08-19T09:23:32+00:00 1 Answers 11 views 0

Answers ( )

  1. Sapo1
    0
    2021-08-19T09:24:34+00:00
    Reply

    Answer:

    6 m/s²

    Explanation:

    Since no net force acts on the box when the velocity is constant, impulse J = Ft = m(v – u) and where F = force acting on box in 0.2 s = 18.0 N, t = time = 0.20 s, u = initial velocity of  box = 0.10 m/s, v = final velocity of box = ? and m = mass of box = 3.0 kg

    So, Ft = m(v – u)

    Since v is unknown, we make v subject of the formula. So,

    v – u = Ft/m

    v = Ft/m + u

    = 18.0 N × 0.2 s/3.0 kg + 0.10 m/s

    = 3.6 Ns/3.0 kg + 0.10 m/s

    = 1.2 m/s + 0.10 m/s

    = 1.3 m/s

    Since v = 1.3 m/s, the acceleration,a of the box produced by the resultant force is

    a = (v – u)/t

    = (1.3 m/s – 0.10 m/s)/0.2 s

    = 1.2 m/s ÷ 0.2 s

    = 6 m/s²

    So, the acceleration produced by the resultant force is 6 m/s²

Leave an answer

Browse

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