A woman pushes a grocery cart filled with groceries up a hill with an 8.0° slope. She stops to rest a distance of 30.0 m up the hill and accidently lets go of the cart. The cart with its groceries has a mass of 18 kg. How long does it take for the cart to roll the distance of 30.0 m down the hill? Ignore friction. 1.6 s, 2.5 s, 6.6 s, or 9.8 s ? How do I solve this? Thanks so much :)
this is physics.
yes:)
Have you learned formulas for distance? It looks like we need one that has mass in it.
sorry it's posted in the maths section! i can't gain access to the physics subject still :(
yes :) but I'm not sure which to use?
is it Fnet=ma ?
Have you seen v = a t \[ s = 1/2 a t^2 \]
looks familiar but i forget how to use it!!
we need distance we need a then we solve for t
distance= 30 a=18 ?
I think a is g, but only the component in the direction parallel to the slope g cos(82)
ohh so how do we sub everything into that equation? :O
apparently we do not need to know the mass (it might matter if we have friction)
a= g cos(82) s= 30
ohh so 30=1/2(cos(82)(t)^2 ?
yes
and then 30=0.474838849(1/2(t^2)) 63.1793293=1/2t^2 ?
oh, don't forget g= 9.8
oh! where do i put that in?
look up above: a= g cos(82)
ohh okay:) so 30= 1/2(9.8*cos(82))(t)^2 30=4.65342072(1/2t^2) 6.44687034=1/2t^2 12.8937407=t^2 t=3.59 ? :/ did i do that right? :/
oh wait oops
oh wait not sure where i went wrong :(
looks like radians instead of degrees
ohhh :( i'm using google as my calculator right now because i left my calculator in my locker at school :( how can i make it degrees on google?
try cos(82 degrees)
okay:)
oh okay yay!! it worked!! so like this then? 30/1.36389639=1/2t^2 21.9958057=1/2t^2 43.9916114=t^2 t=6.6326172 so 6.6 s ? :O
yes, that looks good.
yay!! thanks so much!!! :D will remember to put it on degrees when i use my actual calc!!
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