Static and kinetic friction involving velocity

[enrijuan]

Homework Statement

A baggage handler drops your 7.00 kg suitcase onto a conveyor belt running at 2.30 m/s. The materials are such that µS= 0.410 and µK= 0.250.

Homework Equations

How far is your suitcase dragged before it is riding smoothly on the belt?

The Attempt at a Solution

I know f=µS(Fn) and doesn't apply once sliding occurs. I also know that F=ma.



I really have no idea how to begin.

 

[tiny-tim]

welcome to pf!

hi enrijuan! welcome to pf!

(have a mu: µ )

hint: the variables you have or want are F (force) and d (distance), so you can use … ?

 

[enrijuan]

I know that force over distance is work but I don't see how that would help me in this problem... I am sorry I am having a hard time with this one... I also know that d= vt

 

[tiny-tim]

I know that force over distance is work but I don't see how that would help me in this problem...

i assume you meant force times distance is work?

(never use "over" that way in maths or physics! )

ok, now use the work-energy theorem …

work done = change in mechanical energy ​

 

[enrijuan]

doesnt the conservation of mechanical energy talk about potential and kinetic energy?
I don't understand how I would get potential energy or how work would help me in the problem.

 

[tiny-tim]

doesnt the conservation of mechanical energy talk about potential and kinetic energy?
I don't understand how I would get potential energy or how work would help me in the problem.

yes, mechanical energy is KE plus PE, but the work-energy theorem applies even if the PE is zero (or doesn't change) …

(btw, this isn't conservation of mechanical energy … mechanical energy is lost when there's friction )

in this case, work done = change in KE + change in PE

= change in KE + 0 ​

 

[enrijuan]

so,
work done= (7.00 kg)(2.3 m/s) + 0?

then work done =18.515 J
then 18.515=(force due to kinetic friction)(distance)?

 

[tiny-tim]

so,
work done= (7.00 kg)(2.3 m/s) + 0?

then work done =18.515 J
then 18.515=(force due to kinetic friction)(distance)?

you''re very confused

on the LHS you need work done, which as you say is (force due to kinetic friction)(distance)

(the distance is the unknown that you have to find)

on the RHS you need the change in KE, which is … ?

 

[enrijuan]

oh the end velocity is goin to be zero right?
so,
1/2mv_f²-1/2mv_i²=W

1/2mv_f²-1/2mv_i² = (µmg)(d)

is this right?

 

[enrijuan]

then,
.5(7)(0)-.5(7)(2.3^2)= (.250)(7)(-9.81)d ?

then d=1.08 meters ?

 

[tiny-tim]

oh the end velocity is goin to be zero right?

actually, not in this case …

the question has deliberately put it the other way round, with the initial speed being zero, and the final speed being the same as that of the belt!

(of course, it doesn't make any difference to the size of the change )

so,
1/2mv_f²-1/2mv_i²=W

1/2mv_f²-1/2mv_i² = (µmg)(d)

is this right?

then,
.5(7)(0)-.5(7)(2.3^2)= (.250)(7)(-9.81)d ?

then d=1.08 meters ?

yes!

(though you could have canceled every m in the equation, and saved yourself multiplying by all those 7s! )

 

[enrijuan]

wow thank you so much! thanks for the welcome, the µ, the help, and above all the patience!