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

a

`F = 0.0566 \ N`

b

`t = 6.147 \ s`

Step-by-step 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 + (1)/(2) * a * t^2`

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

So

`1.44= 0 * t + (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 = (F)/(M)`

=>
`a_1 = (0.0566)/(0.656)`

=>
`a_1 = 0.08628 \ m/s^2`

Gnerally from kinematic equation we have

`s = ut + (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 + (1)/(2) * 0.08628* t ^2`

=>
`t = 6.147 \ s`