Question

a 2.34-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 at^2, where a=0.0200 N/s^2

A. What is the speed of the cart 3.5 s after the fan is turned on? (first i got 0.4958 m/s but was wrong)

B. After how many seconds is the cart's velocity zero

Answer 1

(A) The speed of the cart after the fan is turned on which is given as :

The rate of change of velocity (momentum) of an object is proportional to the force :

m v = F t

m (dv / dt) = F

where, F = magnitude of force = a t²

then   m (dv / dt) = a t²

integrating an above eq.

v = a t³ / 3 m                                                                    { eq.1 }

where, a = acceleration = 0.02 N/s²

t = time taken = 3.5 sec

m = mass of cart = 2.34 kg

inserting these values in eq.1,

v = (0.02 N/s²) (3.5 s)³ / 3 (2.34 kg)

v = (0.8575 N.s) / (7.02 kg)

v = 0.122 m/s