Question

A cart on a horizontal, linear track has a fan attached to it. The cart is positioned at one end of the track, and the fan is turned on. Starting from rest, the cart takes 4.22 s to travel a distance of 1.44 m. The mass of the cart plus fan is 350 g. Assume that the cart travels with constant acceleration.
Required:
a. What is the net force exerted on the cart-fan combination?
b. Mass is added to the cart until the total mass of the cart-fan combination is 656 g, and the experiment is repeated. How long does it take for the cart, starting from rest, to travel 1.63 m now?

Answer 1

Answer:

a

  $F = 0.0566 \ N$  

b

   $t = 6.147 \ s$

Explanation:

From the question we are told that

     The distance travel in  4.22 s is  $s = 1.44 \ m$

     The mass of the cart plus the fan is  $m = 350 \ g = 0.35 \ kg$

Generally from kinematic equation we have that

        $s = ut + \frac{1}{2} * a * t^2$

Here  u is the initial  velocity with value  $u = 0 \ m/s$

So  

         $1.44= 0 * t + \frac{1}{2} * a * 4.22^2$      

=>      $a = 0.1617 \ m/s^2$

Generally the net force is  

         $F = m * a$

=>      $F = 0.35 * 0.1617$  

=>      $F = 0.0566 \ N$  

Gnerally the new mass of the cart plus the fan is  $M = 656 \ g = 0.656 \ kg$

    The distance considered is $s_1 = 1.63 \ m$

     Generally the new acceleration of the cart is mathematically represented as

        $F = M * a_1$

=>      $a_1 = \frac{F}{M}$

=>      $a_1 = \frac{0.0566}{0.656}$

=>      $a_1 = 0.08628 \ m/s^2$

Gnerally from kinematic equation we have

          $s = ut + \frac{1}{2} * a_1 * t ^2$

Here u  is the initial velocity and the value is zero because it started from rest  

=>       $1.63 = 0 * t + \frac{1}{2} * 0.08628* t ^2$

=>        $t = 6.147 \ s$