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I don't get this?

  • Thread starter Port
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  • #1
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I need help to figure this problem out. I don't kno what formula to use to solve this!

A shopper in a supermarket pushes a loaded 32kg cart with a horizontal force of 12N. How far will the cart move in 3.5s if the shopper places an 85N child in the cart before pushing.

I only need the formula so plz help!
 

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  • #2
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when they use newtons in terms of weight divide N = mg so 85N/9.8 gives the mass, with that mass add it to the mass of the cart. divide the 12N force by the total mass to find the acceleration. with the acceleration input into formula (Vi)t + .5at^2 = d Vi = zero because the cart is not moving initially therefore velocity = zero. so .5at is left so .5 multiplied by the acceleration you found multiplied by time squared, 3.5s^2
 
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  • #3
dextercioby
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U need to compute the total mass of the cart.Newton's law (theII-nd) says that
[tex] F=ma [/tex]
Once u have found the acceleration,u can use this formula
[tex] s=\frac{1}{2}at^{2} [/tex]

Daniel.
 
  • #4
dextercioby
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Yapper said:
when they use newtons in terms of weight divide N = mg so 85N/9.8 gives the mass, with that mass add it to the mass of the cart. divide the 12N force by the total mass to find the acceleration. with the acceleration input into formula (Vi)t + .5at = d Vi = zero because the cart is not moving initially therefore velocity = zero. so .5at is left so .5 multiplied by the acceleration you found multiplied by time, 3.5s
Wrong.The formula includes a product between the acceleration and the time SQUARED.

Daniel.
 
  • #5
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Yes I corrected it
 
  • #6
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i still don't get it since the answer is supposed to be 2.3M and i get 20m
 
  • #7
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I get 40.7 for total mass 12/40.7(.5)(3.5^2)=1.8M, i get it closer but its still wrong
 
  • #8
dextercioby
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Port said:
i still don't get it since the answer is supposed to be 2.3M and i get 20m
Both answers look wrong to me...What did u get for the mass??What about acceleration??

Daniel.
 
  • #9
dextercioby
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Yapper said:
I get 40.7 for total mass 12/40.7(.5)(3.5^2)=1.8M, i get it closer but its still wrong

I basically did the same calculations.Why one earth would it yield different results??Is there something we missed...?? :confused:

Daniel.
 
  • #10
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for mass i got 40.67kg for mass and 3.39 for acceleration
 
  • #11
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How did u get 3.39? If mass is 40.67 shouldnt acceleration be 12/40.67?
 
  • #12
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how did u guyz get 1.8m i get 20m when i do those calculations
 
  • #13
dextercioby
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Port said:
for mass i got 40.67kg for mass and 3.39 for acceleration
The mass is good,the acceleration is wrong.Wheck your numbers once again.

Daniel.
 
  • #14
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LoL, is that cart driver throwing the child in the cart, thus creating intitial velocity?

Is there anything else to the problem or is this it?

And are you sure the answer is suppsoed to be 2.3M?
 
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  • #15
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ok thats it nvm thanx i kinda also need help on a kinetic friction problem. i am not good with formulas so i don't kno how to solve for this either.

A 4kg block is pushed alomng the ceiling with a constant apllied force of 85N that acts at an angle of 55 degrees with the horizontal. The block accelerates to the right at 6m/s^2. detremine the coefficient of kinetic friction between the block and the ceiling.

the answer is supposed to be 0.816 plz help and thanx again for the last one
 
  • #16
dextercioby
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Do you know the expression for the kinetic friction force??

Daniel.
 
  • #17
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umm i only kno that stuff in the question so i have no clue how to solve
 
  • #18
dextercioby
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Then how would you go about solving it...??I will not solve it for you.It's not the plicy here.We only give indications.
HINT:
[tex] |\vec{F}_{fr.Kin.}|=\mu_{Kin} |\vec{N}| [/tex]

Can u take it from here??
Do u know how to apply Newton's laws??

Daniel.
 
  • #19
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Well you have to find the vertical component of the force using trig, sine will work. That will be your "normal" force. so Fn(CoeFriction)=Ff Ff is the force that friction is exerting in the opposite direction of movment. Take it from there
 
  • #20
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i don't want u guyz to solve it for me of course but i don't understand what either of u are trying to tell me
 
  • #21
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Ok do you have a text book? The teacher should have gave you atleast page numbers to read so you can do the problems. Normal force is the force perpendicular to the surface that the object is on. The product of the Normal force and the coefficient of Kinetic friction gives the force which friction is exerting on the sliding object opposite to the direction of movement. Because the box is being pushed up against the ceiling the ceiling is pushing back with a force(Normal Force), this is equal to the force that the box is being pushed agaisnt the ceiling, which can be calculated with the information in the problem.
 
  • #22
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ok i did the sin thing and i get 69.63N for the normal force. but what do i do after that? iam supposed to solve for the coefficient of friction so how do i get the force of kinetic friction. i am so lost
 
  • #23
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Now find the horizantal compent of the force using cosine. Divide that force by the mass. That will give you the acceleration due to the pushing force, lets call it ap. and lets call the acceleration due to friction af. ap-af=aT, aT being aceleration total. you have aT, 6ms^s, and ap, cause you calculated it using cosine. So now you can calculate af. multiply af by mass to get the force due to friction, Ff. input that into the friction formula and you can find the coeficient of friction by dividing Ff by normal force
 
  • #24
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ok i worked it out but i got the wrong answer. the values i got are: for normal force=69.62N. for ap=12.19. for af=6.19. for Ff=24.8N final answer=.356 and i need .816 what did i do wrong?
 
  • #25
101
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Idk, i get the same answer.

Dex, where are you!!
 

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