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Change in velocity due to force acting over given time

  1. Mar 12, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    What change in velocity would be produced by an unbalanced force of 2.4 × 10^4 N acting for 6 s on a 2.0 × 10^3 kg dragster?

    Not sure how to setup this problem Equations i have are as follows:

    Fnet = mass × acceleration
    Velocity2 = velocity1 + acceleration × time
    Distance = velocity1 × time + 1/2acceleration × time ^2

    d=((v1 + v2) ÷ 2)t
    v2^2=v1^2+2ad

    I don't know how to attempt this question and my teachers hasn't answered my email.

    Please help
     
  2. jcsd
  3. Mar 12, 2016 #2

    Simon Bridge

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    Hint: change in velocity is given by ##\Delta v = v_2-v_1##
    Start by listing what you know: i.e. assign the variables in the equations to the corresponding numbers in the problem.
     
  4. Mar 12, 2016 #3
    Because time 6 s is given only, "Velocity2 = velocity1 + acceleration × time" this formula is enough.
    calculate 'a' first; a = [(2.4 × 10^4 )/(2.0 × 10^3)] = 12 m/s^2
    required change in velocity = (Velocity2 - velocity1) = 12×6 = 72 m/s
     
  5. Mar 12, 2016 #4
    How does 24000N/2000kg = 12 m/s^2?
     
  6. Mar 12, 2016 #5

    Simon Bridge

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    ... hint: do you have an equation with force and mass in it?

    Note: we do not usually do your homework for you.
     
  7. Mar 12, 2016 #6
    Never mind.
     
  8. Mar 12, 2016 #7

    Simon Bridge

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    Do you understand how to do the problem now?
     
  9. Mar 12, 2016 #8

    berkeman

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    Just a reminder -- do not do student's work for them. Please just provide hints, ask probing questions, find mistakes, etc. But it is against the PF homework rules to do their work for them. Makes sense? :smile:
     
  10. Mar 12, 2016 #9
    You could use the equation Impulse = Force x Time to find the change of momentum and subsequently the change in velocity
     
  11. Mar 12, 2016 #10
    I'm so tired I 2000kg as 2kg, and yes understand how to solve the problem.

    Thank you!
     
  12. Mar 12, 2016 #11

    Simon Bridge

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    Nice lateral thinking...
    That approach requires the concept of "specific impulse" and 3 additional equations though, vis: ##\Delta p = m\Delta v## and ##I=Ft## and ##I=\Delta p## ... all that simplifies down to ##F=ma## which OP already has. But sure: that would work and showing various approaches can be helpful.

    ... well done then.

    Just so others benifit more:
    The general approach to a problem like this is to start by drawing a diagram, labelling it, then listing what you know in terms of the symbols you are going to use.
    In the above case you know:
    F=24000N, m=12000kg, and Δt=6s ... you need Δv=? (check the units are consistent)

    Now you want a bunch of equations

    ...by definition Δv=aΔt (constant acceleration) but that has two unknowns.
    You want to find Δv so you need an equation with ##a## in it and F=ma sprang to mind your mind with the others.
    Use algebra to make an equation with what you want on the left and everything you know on the right. In general you may need more than two equations.

    In this case it is a matter of solving the simultaneous equations:
    ##F=ma \implies a=F/m## sub into ##\Delta v = a\Delta t## to give ##\Delta v = F\Delta t /m##

    The last step is plugging the numbers into the equations: ##\Delta v = 24000\times 6 \div 12000 = 72##m/s

    You can also do the number-crunching in two steps, finding the acceleration first.
    In general it is best practise to do all the algebra before you crunch numbers.

    The toughest part of this approach is finding the equations - the usual mistake is to try to do this by memorizing lots of equations. The best practise is to remember the physics that gives rise to the equations. In this case, you need to understand what "acceleration" means. Then writing down the equation is just a matter of using the maths as a language.
     
  13. Mar 13, 2016 #12
    I am replying to the specific sub question of the original asker. How (24000 N)/(2000 kg)
    I hope you know 24000?2000 = 12 Now only you have to expand N in to fundamental units Kg m and s, which comes from the the relation F = ma
    [N] = [(kg×m)/s^2] Hence [N]/[kg] = [m/s^2] You need to repeat this procedure in many other problems to understand it also frame questions your self such as why [J] = [Nm] and also = [(kg m^2)/s^2], Which defining relations in Physics give us these equivalent expressions for [J]. You have a feeling that one has to study and then solve problems. I tell you one more thing study and generate problems to answer.
     
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