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Accelerating shopping cart from rest

  1. Sep 27, 2006 #1
    I'm pretty sure this is an easy problem but I don't know what formula to use -

    At the local grocery store, you push a 16.2 kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N, you now accelerate the cart from rest through a distance of 2.02 m in 3.00 s. What was the mass of the dog food?

    If anyone could just point me in the right direction I would appreciate it...in this class he only throws a bunch of formulas at us and doesn't really explain how to use them so I'm confused.
  2. jcsd
  3. Sep 27, 2006 #2
    F=ma will be the formula you will use in the end. How can you get acceleration from the data you have?
  4. Sep 27, 2006 #3
    So you do F=ma which means a=F/m...so it's 12/16.2 which is .74...right?
  5. Sep 27, 2006 #4
    well that is correct when the cart is pushed with 12 N of force BEFORE the dog food is added. You want to know the mass of the object you are pushing (the cart+dog food) so you would use m=F/a. But you don't have an a value. but you can use a kinematic equation to determine a from the distance and time values you have. [tex] s(t)=\frac{1}{2}at^2 [/tex] should work where s is your distance traveled and t is the time.
  6. Sep 27, 2006 #5
    I'm getting 1.346 for a but if you do m=F/a you get a mass of 8.91 which can't be right because it should be greater than the original 16.2
  7. Sep 27, 2006 #6
    are you remembering to square the t?
  8. Sep 27, 2006 #7
    Yes...s=2.02 and t=3, so I did st=1/2a(t^2) which gave me a=1.346, and since m=F/a I did 12/1.346 but the answer isn't coming out right...I don't know what I'm doing wrong
  9. Sep 27, 2006 #8
    oh, sorry, that was s of t... like f(x)... not s*t. The left side is just s. so its s=1/2a(t^2)
  10. Sep 27, 2006 #9
    ahh, ok...I got it :) thanks!
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