A 2.34-kg cart on a long, level, low-friction track is heading for a small electric fan at 0.24 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 at^2, where a = 0.0200 N/s2.
A) What is the speed of the cart 3.5 s after the fan is turned on?
B) After how many seconds is the cart's velocity zero?
F = ma
Vf = Vi + a*t
The Attempt at a Solution
So for part a I first found the force of the fan on the cart : (0.02)(3.5)^2 = 0.245 N. Then I found the force of the cart, but to do that I first found the acceleration: 0.24 = 0 + a(3.5) where "a" ended up being 0.06857 m/s^2. I plugged this into F = ma to find the force of the cart: F = (2.34)(0.06857) = 0.160457 N. I subtracted the force of the fan from the force of the cart: 0.160457 N - 0.245 N = - 0.0845 N. I then used this value to find the acceleration of the cart with the fan going against it: - 0.0845 N = (2.34 kg)a where "a" ended up being -0.036. I then plugged this back into the kinematic equation: Vf = (0.24 m/s) + (-0.036 m/s^2)(3.5) = 0.11 m/s . Not totally confident in what I did. Could you guide me as to where I went wrong? Thanks!