69 in the Examkrackers physics text book

Here is what I think.
d=tv : this is the average velocity
½mv2 : this is the final velocity
You can't cancel them out like that. They are not equal

 

In the KE equation, it is the speed at any given point you want to determine the energy. In this question, you are asked for the velocity after 2s, so therefore you need the speed at 2s in the equation you chose.

Tieu is right on the mark. The average velocity is half of the final velocity (given that it started at rest): vavg = (vinitial + vfinal)/2 = 1/2 vfinal

If you use your method knowing this, it can work out.

At t = 2:
d = vavg x t = 1/2 vfinal x t

Work = F x d = 100cos30 x 1/2 vfinal x 2 = 87 x vfinal

• KEgained = Workdone on the object
  
  1/2 mvfinal2 = 87 x vfinal
  
  1/2 (10)vfinal2 = 87 x vfinal
  
  5vfinal2 = 87 x vfinal
  
  5vfinal = 87
  
  vfinal = 87/5 = 174/10 = 17.4

You had a good idea. Anytime you are off by exactly a half, you should look back and figure out where the factor comes into play, and often it is because an average value is needed rather than a final value.


So why not try your new and improved method on a variation of that question:

• A 200N force is applied to a 5kg object at an angle of 30 degrees for 3 seconds. If the object is initially at rest, what is its final velocity? (ignore friction: sin 30 degrees= 0.5; cos 30 degrees= 0.87)
  
  A. 34.8 m/s
  B. 52.2 m/s
  C. 104.4 m/s
  D. 208.8 m/s

 

Ever since I was little, I've always done things differently. This works during constant force. You know that F=ma. F= 100cos30=87. 87=F/M=A 87/10=8.7. Acceleration is constant, so after two seconds, it's 8.7*2=17.4. This can be applied to BRT other question as well.

 

A 100N force is applied to a 10kg object at an angle of 30 degrees for 2 seconds. If the object is initially at rest, what is its final velocity? (ignore friction: sin 30 degrees= 0.5; cos 30 degrees= 0.87)
A. 8.7 m/s
B. 1 m/s
C. 17.4 m/s
D. 34.8 m/s

You can also use impulse...

F delta T = m delta v
(100 N*cos30) * 2 secs=10kg (Vf-0)
174=10Vf
Vf=17.4

 

Impulse is change in momentum. Momentum is the product of mass times velocity. Momentum is p=delta mv

Average force x time gives you the change in momentum.
That being said, F delta T = delta mv (or just m delta v if mass doesn't change). It can be derived from Newton's second law F=ma:

F=ma
F=m v/t

Multiple both sides by t, and you get Ft=mv, which is impulse. Learned it in college physics.

 

That is a brilliant solution. Clean and simple and uses all of the information they give you.

Yeah, the answer is C. Your method is just as logical, and not necessarily any longer or shorter than the solution you described from the EK book. Both require two equations. And I'm not sure it's necessarily the EK method, given that kinematics questions were invented at least a hundred years before EK existed.

Now pookiez method, that was a great approach.

And you really didn't encounter impulse when you studied momentum in physics? You can't be serious that it's not in your MCAT review book; you probably just didn't notice it, because it's a major topic. It's pretty heavily tested. You see it in car design and air bag passages. During any collision that transfers momentum, the Faverage x tcollision = delta(mv).

Pookiez's method:
F = 200 N(cos30), t = 3 s, m = 5 kg,
so v = [(200cos30)(3)]/5 = (174 x 3)/5 = 522/5 = 104.4

 

I understand that finding the horizontal force gives the right answer, but why must we find the horizontal force instead of just using the force given (what's the concept/idea behind this)?

Thank you!