1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Force questions?

  1. Jun 22, 2004 #1
    I'm not sure what equation is used on this particular problem.

    How much time would be requird to stop a 2,000 kg car that is moving at 80.0 km/hr if the braking force is 8,000 N?

    I know.....
    Mass 2000 kg
    Acceleration is 80.0 km/hr
    Breaking force is 8,000 N

    What tension must a 50.0 cm length of string support in order to whirl an attached 1,000.0 gram stone in a circular path at 5.00 m/s?

    I used the formula F= mv^2/ r

    = (1,000.0 g) (5.00 m/s)/ 50.0 cm
    = 1,000.0 x 5.00 g*m/s

    I wanted to know if I was tackling this question correctly..if not what am I doing wrong????

  2. jcsd
  3. Jun 22, 2004 #2
    80 km/hr is the velocity, not the acceleration. Just use Newton's second law and some kinematics and you should solve this problem easily.
    You might want to check the units on that again. I suggest you stick to kg, m, and s.
  4. Jun 26, 2004 #3
    terpsgirl...for the first problem....use the equation that states that impulse = change in momentum.....the equation is:
    Force * time = change in velocity * mass.

    Think of what the original velocity is, and the final velocity...and u should be able to find the change in the velocity. Basically plug in your results...and solve for time.

    for the second problem.....i would check ur units....notice what unit ur expressing r in....and look at the units ur measuring speed in. Remeber that u should measure everything in kg's, meters and seconds.
    Last edited: Jun 26, 2004
  5. Jun 26, 2004 #4
    First convert to the correct units.
    Car = 2000kg.
    Speed = 22.22... m/s.
    Braking force = 8000N.

    Using F=MA. 8000=2000A, the 8000N force acting against the car will create a deceleration of magnitude 8000/2000, thus, a = -4m/s/s.

    U = 22.22....., V = 0, a = -4, t = ?
    V = u + at
    0 = (200/9) + (-4t)
    (200/9) = 4t
    t = 5, 5/9 seconds
    = 5.56 seconds (2dp)

    I haven't studied the concept used in question 2, but the equation looks like it would be effective. However, [always] convert to more useful units. You should be working in m/s, kgs, Ns and M.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook