1
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?

1 Answer

2

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

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Stevie Lemke
15.5k 3 10 26
answered 11 months ago