UFPM Blog

Unit Summary UFPM

This unit covered a vast amount of information in regards to unbalanced forces but was also really exciting. Examples of such forces are all balls in the air, as only gravity acts upon them, all accelerating objects, and anything really that is not maintaining a constant velocity.

The formula’s used were:

1.       vf = at + vi

2.       vf² = vi² + 2aΔx

3.       a = F_net/mass (in kg)

4.       Δx = 1/2 aΔt² + ViΔt

5.       a = Δv/Δt

We started this unit exploring what happens when there is more force going one way, than holding it back, by looking at what happened if friction, or air resistance were not factored in, or if they were, that they were not equal in force.

Examples of such scenarios are cars that are speeding up or slowing down. Part of this was learning how to solve for the frictional force, as shown below.

If a car that weighs 240 kg with an F_push  of 300n, and and F_F of 80, the following diagram looks like:

In order to solve for acceleration, we must use the a=fnet/m formula as shown below. 

We have solved for acceleration, it is .91m/s^2

If we know that a parachutist covered 1500 meters before expanding his parachute, from rest, how long did it take him?

                     It took him 17.3 seconds to fall 1500 meters.

Now, what is his final velocity right before he opens his parachute?

Thus it is 173m/s before he pulls his parachute cord open. 

This is the summary of our blog, if you have questions, please feel free to comment below.