# 69 in the Examkrackers physics text book

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#### [email protected]

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Please refer to number 69 in the Examkrackers physics text book (it has the diagram which may be useful).

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

I solved this equation using the W=K. I figured that there was no friction so there was no heat and no change in internal energy. The object is also on a flat surface with no incline so there is no potential energy to worry about and all the energy is transferred into Kinetic Energy.
So W=K, W=Fdcos&#952;, K= ½mv2
Thus, Fdcos&#952;= ½mv2
F(tv)cos&#952;= ½mv2 simplified is F(t)cos&#952;= ½mv
Now we can plug and chug F=100N, t=2s, &#952;=30 degrees, m=10kg, and we are solving for v
(100N)(2s)cos30= ½(10kg)v
v=34.8 m/s which is answer d

Except, according to exam krackers the answer is c. Their explanation is The horizontal component of the force is 87N. To find the acceleration we use F=ma. a=8.7 m/s2. We use v=vo+at and arrive at 17.4 m/s.

I understand how they solved it. But why is the way I solved it wrong. What did I do wrong? Can someone explain this to me.

Thank you,

Veronica

#### TieuBachHo

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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

#### [email protected]

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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

I guess that makes sense. Where does it say that the v in K=1/2mv2 is final velocity? It is not in the EK book or in any text book that I have or on-line.

#### BerkReviewTeach

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I guess that makes sense. Where does it say that the v in K=1/2mv2 is final velocity? It is not in the EK book or in any text book that I have or on-line.

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

#### TheBoondocks

##### StreetFighter 4 Virtuoso
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Please refer to number 69 in the Examkrackers physics text book (it has the diagram which may be useful).

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

I solved this equation using the W=K. I figured that there was no friction so there was no heat and no change in internal energy. The object is also on a flat surface with no incline so there is no potential energy to worry about and all the energy is transferred into Kinetic Energy.
So W=K, W=Fdcos&#952;, K= ½mv2
Thus, Fdcos&#952;= ½mv2
F(tv)cos&#952;= ½mv2 simplified is F(t)cos&#952;= ½mv
Now we can plug and chug F=100N, t=2s, &#952;=30 degrees, m=10kg, and we are solving for v
(100N)(2s)cos30= ½(10kg)v
v=34.8 m/s which is answer d

Except, according to exam krackers the answer is c. Their explanation is "The horizontal component of the force is 87N. To find the acceleration we use F=ma. a=8.7 m/s2. We use v=vo+at and arrive at 17.4 m/s.

I understand how they solved it. But why is the way I solved it wrong. What did I do wrong? Can someone explain this to me.

Thank you,

Veronica

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.

#### [email protected]

##### Member
15+ Year Member
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.

Wow thanks! Your way is the simplest way I have seen. i did mine out real complexly.

#### G1SG2

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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

#### [email protected]

##### Member
15+ Year Member
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

Is the answer C? I used EK method for this since there's was simpler. I seem to come up with more complex ways to solve problems. Which I hope does not eat up my time on the real MCAT.

#### [email protected]

##### Member
15+ Year Member
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

What is impluse? Where did u learn about that? I have never seen that before.

#### G1SG2

##### Full Member
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What is impluse? Where did u learn about that? I have never seen that before.

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.

#### BerkReviewTeach

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You can also use impulse...

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

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

Is the answer C? I used EK method for this since there's was simpler. I seem to come up with more complex ways to solve problems. Which I hope does not eat up my time on the real MCAT.

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

#### [email protected]

##### Member
15+ Year Member
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 don't remember learning it in Physics which was 4 years ago now or in my Kaplan physics MCAT class. I have not gotten to the momentum chapter in my EK Physics books but now that I look it is there. Thank you for all your help!

Veronica

ps. If you have time.... I have been having issues with some torque problems. Do you think you might be able to help me out?

#### Rapha

##### Full Member
5+ Year Member
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! 