Answer :
Answer:
a
[tex]F = 0.0566 \ N[/tex]
b
[tex]t = 6.147 \ s[/tex]
Explanation:
From the question we are told that
The distance travel in 4.22 s is [tex]s = 1.44 \ m[/tex]
The mass of the cart plus the fan is [tex]m = 350 \ g = 0.35 \ kg[/tex]
Generally from kinematic equation we have that
[tex]s = ut + \frac{1}{2} * a * t^2[/tex]
Here u is the initial velocity with value [tex]u = 0 \ m/s[/tex]
So
[tex]1.44= 0 * t + \frac{1}{2} * a * 4.22^2[/tex]
=> [tex]a = 0.1617 \ m/s^2[/tex]
Generally the net force is
[tex]F = m * a[/tex]
=> [tex]F = 0.35 * 0.1617[/tex]
=> [tex]F = 0.0566 \ N[/tex]
Gnerally the new mass of the cart plus the fan is [tex]M = 656 \ g = 0.656 \ kg[/tex]
The distance considered is [tex]s_1 = 1.63 \ m[/tex]
Generally the new acceleration of the cart is mathematically represented as
[tex]F = M * a_1[/tex]
=> [tex]a_1 = \frac{F}{M}[/tex]
=> [tex]a_1 = \frac{0.0566}{0.656}[/tex]
=> [tex]a_1 = 0.08628 \ m/s^2[/tex]
Gnerally from kinematic equation we have
[tex]s = ut + \frac{1}{2} * a_1 * t ^2[/tex]
Here u is the initial velocity and the value is zero because it started from rest
=> [tex]1.63 = 0 * t + \frac{1}{2} * 0.08628* t ^2[/tex]
=> [tex]t = 6.147 \ s[/tex]