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)?

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moya1316 moya1316 · Answer 1 · 2019-11-18T21:29:54+01:00

Answer:

[tex]v_{f}[/tex] = 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 [tex]v_{f}[/tex] - m v₀

I = m [tex]v_{f}[/tex]- 0

[tex]v_{f}[/tex] = I / m

[tex]v_{f}[/tex] = 5350.7 / 1.42

[tex]v_{f}[/tex] = 3768 m / ms

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