A Truck Braking

  • Thread starter postfan
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  • #1
postfan
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Homework Statement


A truck is initially moving at velocity v. The driver presses the brake in order to slow the truck to a stop. The
brake applies a constant force F to the truck. The truck rolls a distance x before coming to a stop, and the time
it takes to stop is t.
8. Which of the following expressions is equal the initial kinetic energy of the truck (i.e. the kinetic energy before the driver starts braking)?
(A) F x
(B) F vt
(C) F xt
(D) F t
(E) Both (a) and (b) are correct
9. Which of the following expressions is equal the initial momentum of the truck (i.e. the momentum before the driver starts braking)?
(A) F x
(B) F t/2
(C) F xt
(D) 2F t
(E) 2F x/v

Homework Equations


W = delta K
p= m*v
F*t = delta p

The Attempt at a Solution


For 8 I used the work energy theorem to get the kinetic energy to equal Fx. And since x=vt, I picked (E) but apparently (A) is right. Why doesn't x=vt apply?
For 9 I used the fact that force* time = delta p, to get that p=F*t, and that's where I get stuck. Any thoughts on how to continue?
 

Answers and Replies

  • #2
arpon
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The final velocity of the truck is 0 . So, the final momentum is also 0 . And ##\Delta p = p_{final} - p_{initial} = 0 - p_{initial} = - p_{initial}## . Here, the sign is negative because the force is applied opposite to the direction of the initial velocity of the truck.
And look, ##x = (\frac{v_{initial} + v_{final}}{2}) \cdot t ##
 
Last edited:
  • #3
Nathanael
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For 8 ... Why doesn't x=vt apply?
The velocity is not constant.

For 9 I used the fact that force* time = delta p, to get that p=F*t, and that's where I get stuck. Any thoughts on how to continue?
Which answers can you eliminate?
 
  • #4
postfan
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Oh ok I get 8 then, basically x=vt doesn't apply when accelerating. OK.
As for 9 I can eliminate answers A and C from dimensional analysis.
 
  • #5
Nathanael
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Oh ok I get 8 then, basically x=vt doesn't apply when accelerating.
8.(B) would be correct if v was the average velocity instead of the initial velocity.

As for 9 I can eliminate answers A and C from dimensional analysis.
Ok good. You said you think p=F*t, so 9.(B) and 9.(D) don't agree.
What about 9.(E), that seems like a strange one, what are your thoughts on the meaning of that answer?
 
  • #6
postfan
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I would have said that they're using t=x/v but since we have established that only works for average velocity, I don't really know.
 
  • #7
Nathanael
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I would have said that they're using t=x/v but since we have established that only works for average velocity, I don't really know.
Right, but we know that the force is constant. Can you use that to determine what the average velocity is?
 
  • #8
postfan
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Since there is constant acceleration due to a constant force and the final velocity is 0, isn't the average v just equal to v_initial/2?
 
  • #9
Nathanael
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Good. So you said t=x/vavg and you said vavg=vi/2, then what is x/vi?
 
  • #10
postfan
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It is equal to t/2.
 
  • #11
Nathanael
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And so 9.(E) is....?

(So many problems at once! :) )
 
  • #12
postfan
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Since p=Ft and x/v_i=t/2 then t equals 2x/v_i and the answer is just p=2x/v_i. But doesn't (E) refer to the initial velocity and not the average velocity?
 
  • #13
Nathanael
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Since p=Ft and x/v_i=t/2 then t equals 2x/v_i and the answer is just p=2x/v_i
I think you mean p=F*(2x/vi) ?

But doesn't (E) refer to the initial velocity and not the average velocity?
Right. Aren't we referring to the initial velocity when we say vi?
 
  • #14
postfan
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Ok yes to both counts, just out of curiosity if Ft was an answer choice would that be right?
 
  • #15
Nathanael
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Yes Ft would also be right, but I guess they didn't want to make it too easy :)
(At least they didn't give an option of "none of the above," that would probably trick a number of students!)

2Fvi/x is just a strange way of writing Ft
 
  • #16
postfan
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Ok well the Physics Olympiad (where I got the questions) loves to make things difficult. Anyways, thanks for your help!
 

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