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.

F = .0566 N still

m = .733
a = F/m = .0772
1.44 = .5 (.0772) t^2
t = 6.11 s

Wow it was that easy. Thank You!!!

To approach question 2, you can use the equation of motion:

d = (1/2) a t^2

where d is the distance traveled, a is the acceleration, and t is the time.

Given that the distance (d) is 1.44 m and you want to solve for the time (t), you can rearrange the equation as follows:

t^2 = (2d) / a

Now, you can substitute the values of d and a from the previous problem into the equation:

t^2 = (2 * 1.44) / 0.1617

Simplifying, you get:

t^2 = 17.69

To solve for t, you take the square root of both sides:

t = √17.69

Calculating that, you get:

t ≈ 4.2 s

Therefore, when the mass of the cart-fan combination is increased to 733 g, it takes about 4.2 seconds for the cart to travel 1.44 m starting from rest.

To solve question 2, we can use the concept of Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

First, we can calculate the acceleration of the cart-fan combination in question 1 using the formula:

a = (2d) / t^2

Using the given values, we can substitute them into the formula to find the acceleration in question 1:
a1 = (2 * 1.44) / (4.22^2)
= 0.1617 m/s^2

Now, let's calculate the force exerted on the cart-fan combination in question 1 using Newton's second law of motion:

F1 = m1 * a1

Substituting the known values, we have:
F1 = (0.35 * 0.1617)
= 0.0566 N

In question 2, the total mass of the cart-fan combination is given as 733 g. Let's convert this mass to kilograms:
m2 = 733 g / 1000
= 0.733 kg

Using Newton's second law, we can find the acceleration of the cart-fan combination in question 2:

a2 = F2 / m2

Since the force remains the same as in question 1, we can use the same force value from question 1:
a2 = F1 / m2
= 0.0566 / 0.733
≈ 0.0772 m/s^2

Now, we can use the formula of motion to find the time it takes for the cart, starting from rest, to travel a distance of 1.44 m in question 2. According to the formula:

d = (1/2) * a2 * t^2

Rearranging the formula, we get:

t^2 = 2d / a2

Substituting the known values:
t^2 = (2 * 1.44) / 0.0772
t^2 ≈ 37.3
t ≈ √(37.3)
≈ 6.1 s

Therefore, it will take approximately 6.1 seconds for the cart, starting from rest, to travel a distance of 1.44 m in question 2 with the increased mass.