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.

Asked 11 months ago

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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?

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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

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

=> $t = 6.147 \ s$

Edit

Stevie Lemke

15.5k 3 10 26

answered 11 months ago