Question

A stationary 1.42 kg object is struck by a stick. The object experiences a horizontal force given by F = at - bt3 , where t is the time in milliseconds from the instant the stick first contacts the object. If a = 1500 N/ms and b = 20 N/ms3, what is the speed of the object just after it comes away from the stick (t = 2.74 ms)?

Answer 1

Answer:

 $v_{f}$ = 3768 m/ms

Explanation:

For this exercise we will use the relationships of the moment

    I = ∫ F dt = Δp

The expression for strength is

    F = at - bt3

Let's replace and calculate the momentum

    I = ∫ (at - bt3) dt

    I = a t2 / 2 - b t4 / 4

We evaluate between the initial time of zero and the final time of 2.74 ms

    I = ½ a (2.74)² - ¼ b (2.74)⁴

    I = ½ 1500 7.51 - ¼ 20 56.36

    I = 5632.5-281.8

    I = 5350.7 N ms

Now we use the momentum relationship with the moment

    I = Δp

    I = m $v_{f}$ - m v₀

    I = m $v_{f}$- 0

   $v_{f}$ = I / m

   $v_{f}$ = 5350.7 / 1.42

  $v_{f}$ = 3768 m / ms

   $v_{f}$ = 3768 m / ms (1000 ms / 1 s) = 3.768 106 m / s