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A cart on a horizontal, linear track has a fan attached to it.The cart is positi

ID: 1764299 • Letter: A

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 isturned on. Starting from rest, the cart takes 4.45 s to travel adistance of 1.35 m. The mass of the cart plus fan is 385 g. Assumethat the cart travels with constant acceleration. (a) What is the net force exerted on the cart-fancombination? ...... N
(b) Mass is added to the cart until the total mass of thecart-fan combination is 649 g, and the experiment is repeated. Howlong does it take for the cart, starting from rest, to travel 1.35m now? Ignore the effects due to friction. .......s 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 isturned on. Starting from rest, the cart takes 4.45 s to travel adistance of 1.35 m. The mass of the cart plus fan is 385 g. Assumethat the cart travels with constant acceleration. (a) What is the net force exerted on the cart-fancombination? ...... N
(b) Mass is added to the cart until the total mass of thecart-fan combination is 649 g, and the experiment is repeated. Howlong does it take for the cart, starting from rest, to travel 1.35m now? Ignore the effects due to friction. .......s

Explanation / Answer

       (a) Given that the length oftrack is S = 1.35 m             Time taken is t = 4.45s             Mass of cart and fan is m = 0.385kg       From the equationof motion is S = ut + (1/2)at2                                                                    1.35m = 0 + (1/2)a(4.45s)2                                                                                       a = 0.136 m/s2    Net force is F =ma                                  = 0.0525N (b) If mass of cart is m = 649 g .    Since the force issame F = 0.0525 N    Then accelerationis a = F / m                                              = 0.0525N/ 0.649kg                                              = 0.08m/s2                     Then distance travel is S = ut + (1/2)at2                                        1.35m = 0+ (1/2)(0.08m/s2)t2                                                   t = 5.80 s

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