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HELP WITH AP PHYSICS- 5 min drill! Multiple choice easy?

  1. Feb 5, 2012 #1
    I need help with all of these. I am so lost on all of them and have been working for hours. Anything is so appreciated.

    _____1. A rock is dropped off a cliff and falls the first half of the distance to the ground in t1 seconds. It falls the second half of the distance in t2 seconds. What is the value of t1/t2?
    A) 1/2√2 B) 1/√2 C) ½
    D) 1-(1/√2) E) √2-1

    ______2. A box of mass m slides on a horizontal surface with initial speed v0. It feels no forces other than it weight and the force from the surface. If the coefficient of kinetic friction between the box and surface is μ, how far does the box slide before coming to rest?
    A) v02/2μg B) v02/μg C) 2v02/μg
    D) mv02/μg E) 2 mv02/μg

    _____3. An object initially at rest experiences a time-varying acceleration given by a = ( 2 m/s3 t) for t ≥ 0. How far does the oject travel in the first 3 seconds?
    A) 9 m B) 12 m C) 18 m D) 24 m E) 27 m

    ______4. An engine provides 10kW of power to lift a heavy load at constant velocity a distance of 20 meters in 5 seconds. What is the mass of the load?
    A) 100 kg B) 150 kg C) 200 kg D) 250 kg E)300 kg

    ______5. A ball (mass= .08 kg) is dropped onto the floor from a height of 2 m, and after bouncing off the floor rises almost to its original height. If the contact time with the floor was .04 s, what average force did the floor exert on the ball?
    A) 0.16 N B) 16 N C) 25 N D) 36 N E) 64 N
  2. jcsd
  3. Feb 5, 2012 #2
    For #4, I said Work/Time=Power so Fd/t=P so Ma*d/t=P because W=fd=ma*d so then I plugged in all the numbers and got .25 which leads me to believe its 250. ??? I dont know I picked D
  4. Feb 5, 2012 #3


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    Welcome to PF girlphysics!

    For this first problem, you can just use the relevant kinematics equation. What you want is the equation that tells you how distance varies with time for motion with constant acceleration. Start with that. Do you see any equations from your notes or your book that give distance vs. time for the case of a constant acceleration?
  5. Feb 5, 2012 #4
    v0t+1/2at^2 so because initial is 0, just 1/2at^2??
  6. Feb 5, 2012 #5
    How do you suppose you would find out what t1 is?
    Since I see you know what work is, how much work must be done on the box to make it come to rest?
    Do you know any calculus?
    You may be missing a factor of 1000 because the question uses kW = 1000 W
    What is the definition of force?
  7. Feb 5, 2012 #6


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    Your understanding of the physics seems fine. power = work/time = force*distance/time. Hence: P = mg*d/t

    Solve for m.

    I suspect that you simply made a mistake with the units. Be careful: that's 10 kilowatts, not 10 watts.
  8. Feb 5, 2012 #7
    Let me just say, I have a terrible teacher and probably shouldnt be in AP physics, its halfway through the year and I know nothing. THANK YOU for your help. I also have not taken calculus and I have to turn this in tomorrow.
    1.) t=√1/2ax ? Im not sure. I feel really stupid
  9. Feb 5, 2012 #8


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    Exactly. Now what is "a" in this situation? Hint: what is the only force that is present here?

    So, you can solve for t1 using this equation. Call the initial height above the ground "h".


    h/2 = 1/2*(a)*t12

    Solving for t2 is trickier. You consider the halfway point to be the start of a new problem where you apply the same equation again. Only now there IS an initial velocity: it's equat to the velocity the object has reached at the half-way point. To compute this, you need another kinematics equation that gives you velocity vs. time.
  10. Feb 5, 2012 #9
    OK So I got 4! yay. 1 down 4 to go.
  11. Feb 5, 2012 #10
    a is 10m/s/s gravity...
    so h= 5t^2 so t=√h/5
  12. Feb 5, 2012 #11
    I dont know what equation..?
  13. Feb 5, 2012 #12
    2.) I dont know it must only be effected by the normal force and coefficient of friction, and F=μFn. I dont know what to do.
    5.) F=ma. I worked on this for like an hour. Dont I need the change in velocity? to find the impulse?
  14. Feb 5, 2012 #13
    I just need the letter answers if its easier to work backwards...
  15. Feb 5, 2012 #14


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    Yeah exactly. The answer I was going for was that a = g, where g is defined as 9.81 m/s/s (or 10 m/s/s if you're not too concerned about accuracy).

    Your work is pretty close. Be a little more careful with your algebra. The equation was

    h/2 = (1/2)gt12

    multiply both sides by 2:

    h = gt12

    t1 = √(h/g)

    You don't know how velocity varies with time during constant acceleration? Hint: what is the definition of acceleration?

    The question you're trying to answer is: how fast is it going when it reaches the halfway point?
  16. Feb 5, 2012 #15


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    The wording of Question #1. must be in error. The time, t1 to go the first half of the distance must be greater than time, t2, the time to go the second half of the distance.

    Therefore t1/t2 > 1 . But ALL of the answers are less than 1.

    Perhaps if the time t2 represented the time to go the total distance, then one of the listed answers might be correct.
  17. Feb 5, 2012 #16
    oh that makes sense!

    well acceleration is the rate at which velocity changes...

  18. Feb 5, 2012 #17
    but theres also v^2=v0^2+a(x-x0)
  19. Feb 5, 2012 #18
    THATS exactly what I thought at first, but I know this question is right....
  20. Feb 5, 2012 #19


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    It's actually v = v0 + at. (Remember that equations have to be dimensionally consistent: if the right-hand side has dimensions of velocity, the left-hand side must as well, otherwise it is nonsense).

    Can you see how to use this equation (velocity vs. time) to figure out how fast the object was going at time t = t1 when it reaches the half-way point?

    How do you know? It could easily have a typo...
  21. Feb 5, 2012 #20
    Well plug in t1 for t?
    I dont know for sure, but I asked two friends who have the same problem if they were able to find the right answer and they both said yes, and Im pretty sure my teacher has been using the same problems for like his 30 years in teaching- not saying that everyone could have missed the typo.
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