Solving Newton's Laws Problem: Stopping a Cart w/ Box in Min Time

[sedaw]

A box put on a cart with velocity of 40m/s.
The static drag coefficient between the box and the cart is 0.3.
How long (in metes) before the cart stop should the driver start to slow down if he want to stop the cart on minimal time and make the box not move ?

im received 544.2 m but the answer is 227 .

tnx.

 

[berkeman]

A box put on a cart with velocity of 40m/s.
The static drag coefficient between the box and the cart is 0.3.
How long (in metes) before the cart stop should the driver start to slow down if he want to stop the cart on minimal time and make the box not move ?

im received 544.2 m but the answer is 227 .

tnx.

Please show us your work -- how did you arrive at 544.2m as your answer?

 

[sedaw]

sigmaFy=0 ----> mg=N
sigmaFx=ma --->a=-f_s/m

note: f_s = static friction

f_k=y*N = y*mg

recived : a= -yg = -0.3*9.8 ====> a = -2.94m/s^2


vo*t +0.5at^2=x

x will be 0 cause the box not move

X=0
v0=40
a=-2.94
t= ?


received t=27.21s


what`s the way will past ?

vt^2=vo^2=2ax
vt will be 0 cause the cat stop after t seconds.

v0=40
vt=0
t=27.21
x=?

received X=544.2


wts wrong ?

 

[LowlyPion]

sigmaFy=0 ----> mg=N
sigmaFx=ma --->a=-f_s/m

note: f_s = static friction

f_k=y*N = y*mg

recived : a= -yg = -0.3*9.8 ====> a = -2.94m/s^2vo*t +0.5at^2=x

x will be 0 cause the box not move

X=0
v0=40
a=-2.94
t= ?received t=27.21swhat`s the way will past ?

vt^2=vo^2=2ax
vt will be 0 cause the cat stop after t seconds.

v0=40
vt=0
t=27.21
x=?

received X=544.2wts wrong ?

Isn't the time to stop going to be given by how long at constant deceleration it takes to get to 0? From 40m/s and decelerating at 2.94 m/s isn't that just 40/2.94 for the time?

Using the time you can then determine the distance with 1/2 a t²

 

[sedaw]

the deceleration constant .

forgat to mention .

there`s any offer ?

 

[sedaw]

Using the time you can then determine the distance with 1/2 a t²

u mean that : vo*t+0.5at^2=X ?

 

[LowlyPion]

u mean that : vo*t+0.5at^2=X ?

Yes, but I was simplifying it by observing that the final velocity for deceleration was going to be 0 and hence could be represented by just the simple x = 1/2 a t².

If you prefer you can use the relationship that

$$v^2 = v_0^2 + 2 a \Delta x$$

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

Here the final velocity is 0 so that makes it merely (40)²/(2*2.94) = X

 

[sedaw]

first for determine distance i can use (V0+Vt)*t/2

but how i can be sure that the not move while the cart decelerate ?

TNX ...

 

[PhanthomJay]

You determined the max acceleration with respect to the ground of the box based on the max available static friction force. Then you set the acceleration of the cart with respect to the ground the same as the acceleration of the box with respect to the ground. Is the box moving with respect to the cart?

 

[LowlyPion]

first for determine distance i can use (V0+Vt)*t/2

but how i can be sure that the not move while the cart decelerate ?

TNX ...

OK you can use that too. The right answer is the same no matter how you get there.

You know the box is not going to move because you already determined the maximum deceleration the box could tolerate while slowing without it moving.