Question

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.

Answer 1

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

Step-by-step 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

`a=(F)/(m)`

`a=(at^2)/(m)`

The speed of the cart at time 3.5 sec is

Using equation of motion

`v = u+at`

Put the value into the formula

`v=0+(at^2)/(m)* t`

`v=(0.0200)/(2.29)*(3.5)^3`

`v=0.37\ m/s`

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