UFPM: Nikki and Elise

Materials:

-cart

-buggy

-pulley

-Atwood machine

-500 kg weight

First we took the mass in grams of all variables in the experiment, and converted them to Newtons.

In this experiment, the goal was to predict the point that the buggy and the pulley would intersect. To make this prediction, we had to find the following variables:

•   net force

•   velocity of the buggy

mass and net force on the Atwood Machine

In this experiment, we did not take the acceleration of the buggy became the force was balanced since it moved at constant velocity.

Item	Mass (kg)
Buggy	.477
Cart	.9892
Pulley	.0509
Total system	1.04

First, we plugged in Fg, which was -10.04N, and could conclude that the diagram was balanced in the vertical direction, so therefore Fn was also 10.4N.  We knew that it was frictionless along the surface, so we concluded friction was 0N. Next, to solve for Ft, we used the equation: Fnet= -Ff+ Ft (which is equal to ma).

Using the data we found for the mass of the pulley, we found the force of this mass due to gravity.

.0509 kg

Fg2= 0.509N

After I found the Fnet in the horizontal direction, I plugged it and the mass into A=Fnet.

              Mass

Then I used the equation for displacement to solve for time of the hanging mass to drop.

ΔX= 1/2at^2+Vit

.86 = 1/2at^2 + Vit

(2) .86 = .5 t^2

1.8547 = t

Prediction:

1.85s =t

I also timed how long it would go for the buggy to go one meter. Since it is moving at constant speed, acceleration is equal to 0. So the equation becomes:

ΔX= Vit

Vi= ΔX  = 1 m= 0.25 m/s

       T         4

Now that we found the velocity, we can use the time it took for the hanging mass to drop and solve for the displacement of the buggy.

Δx= Vit

(.25)(1.85)

ΔX= .4625m

=46.25cm back from where the pulley will land.

Our pulley landed on the cart, so the experiment had 0% error.