Refer to procedure 2. Assume the mass of the cart is m , and the mass hanging on

Question

Refer to procedure 2. Assume the mass of the cart is m, and the mass hanging on the end of the string is M.

a) Suppose m = 1821 g and M = 648 g, and the cart accelerates at 0.418 m/s2. Find ?k, the coefficient of kinetic friction between the cart and the track.


b)Suppose you attached two identical carts of mass m = 1821 g to one end of the string, and doubled the mass M = 648 g. Also, assume the coefficient of kinetic friction between each cart and the surface is the same as in part a. When this system accelerates, the acceleration will be

Explanation / Answer

(a) Let tension in wire be T.

The acceleration of cart m = 1.821 kg and mass M = 0.648 kg are same. Let it be a, Then a = 0.418 m/s2

Then, Mg - T = Ma and T - kmg = ma (Since, the normal reaction force on cart is N = mg, and kinetic friction force = kN)

Eliminating T, we have: Mg - Ma - kmg = ma

So, k = (Mg - Ma - ma)/mg = (0.648*9.8 - 0.648*0.418 - 1.821*0.418)/(1.821*9.8) = 0.298

(b) Here we have cart of mass 2m, and hanging object of mass 2M. Also, k =0.298.

Note that 2Mg - 2Ma' - 2kmg = 2ma' => a' = (2Mg - 2kmg)/(2M + 2m) = (Mg - kmg)/(M + m) = a, that is, acceleration will be same as part (a).

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