1. Not finding help here? Sign up for a free 30min 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!

Hw problem

  1. Oct 30, 2003 #1
    The questin is a 5kg box slides down a frictionless plane. The plane makes an angle of 15 degrees with the horizonal and has a length of 2.00 meters.
    a) if the boxs starts from rest calculate the acceleration
    b) calculate the speed when the box reaches the bottom of the incline
    I did draw a picture.

    I have w = mg which is w 5(9.80) = 49 N
    Wx = sin 15 *49 = 12.68 N Wy = cos 15 * 49 = 47.33 N

    To determine the accel i used the formula A=F/m
    A= 12.68 N /5kg = 2.53m/s

    D = 1/2 AT^2 2*2.53*t^2 = 1.26 secoinds

    Vf^2 = Vi^2+2AD =Vf^2 = 2*2.53* 2 by the way Vi = 0 since it started from rest
    Vf = 3.18m/s

    Is this right?
  2. jcsd
  3. Oct 30, 2003 #2


    User Avatar
    Science Advisor
    Gold Member

    The method is correct (I'm sure the numbers are too, but I didn't check them). Not to split hairs, but when you label acceleration, watch your units: it should be in m/s/s for this problem. (People have been known to lose points on hws and quizzes for such things.)
  4. Oct 31, 2003 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    You might want to specify what "x" and "y" mean. It appears to me that you are taking your x axis along the incline and y axis normal to it. Is that what you intended? (Actually, I'm sure it is and you have calculated the acceleration correctly.)

    You don't really need to calculate the weight by multiplying by the mass and then dividing by the mass to get acceleration- as in all "gravitational" problems (without friction), the mass is irrelevant. The acceleration along the incline is 9.8 sin(15)= 2.53 m/s2 just as you got (but NOT "m/s" as you write!), though I think I did less work to get it!

    Finally, your equation "D = 1/2 AT^2 2*2.53*t^2 = 1.26 secoinds" is mystifying! No, D (a distance) is not 1.26 seconds! I THINK what you did is solve the equation
    (1/2) (2.53)T^2= 2 to get T= &radic:(2*2/2.53)= 1.26 seconds, but that is certainly not what you said!

    Now that you know the time the box was sliding and its acceleration,
    v= aT= (2.53 m/s2)(1.26 s)= 3.18 m/s.

    You were right! (Which, considering how you wrote this out, is something of a miracle!)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Hw problem
  1. HW problem (Replies: 9)

  2. Hw problems (Replies: 3)

  3. HW Problem (Replies: 3)