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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!

- Thread starter Port
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- #1

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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!

- #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

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[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

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Wrong.The formula includes a product between the acceleration and the time SQUARED.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

Daniel.

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Yes I corrected it

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i still don't get it since the answer is supposed to be 2.3M and i get 20m

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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

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Both answers look wrong to me...What did u get for the mass??What about acceleration??Port said:i still don't get it since the answer is supposed to be 2.3M and i get 20m

Daniel.

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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...??

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

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The mass is good,the acceleration is wrong.Wheck your numbers once again.Port said:for mass i got 40.67kg for mass and 3.39 for acceleration

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?

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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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

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Do you know the expression for the kinetic friction force??

Daniel.

Daniel.

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umm i only kno that stuff in the question so i have no clue how to solve

- #18

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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.

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Idk, i get the same answer.

Dex, where are you!!

Dex, where are you!!

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