A 2.29-kg cart on a long, level, low-friction track is heading for a small electric fan at 0.21 m/s . The fan, which was initially off, is turned on. As the fan speeds up, the magnitude of the force it exerts on the cart is given by at2, where a = 0.0200 N/s2.

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CarliReifsteck CarliReifsteck · Answer 1 · 2019-10-18T11:45:55+02:00

Answer:

The speed of the cart at time 3.5 sec is 0.37 m/s.

Explanation:

Given that,

Mass = 2.29 kg

Speed = 0.21 m/s

Force = at²

a = 0.0200 N/s²

Suppose we find the speed of the cart at 3.5 sec after the fun is turn on

We need to calculate the acceleration

Using newton's law

[tex]a=\dfrac{F}{m}[/tex]

[tex]a=\dfrac{at^2}{m}[/tex]

The speed of the cart at time 3.5 sec is

Using equation of motion

[tex]v = u+at[/tex]

Put the value into the formula

[tex]v=0+\dfrac{at^2}{m}\times t[/tex]

[tex]v=\dfrac{0.0200}{2.29}\times(3.5)^3[/tex]

[tex]v=0.37\ m/s[/tex]

Hence, The speed of the cart at time 3.5 sec is 0.37 m/s.