Distance Calculations for Sliding Box with Given Forces

[mortho]

[SOLVED] Finding distance using forces

Homework Statement

A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.18 and the push imparts an initial speed of 4.2 m/s?

Homework Equations

x=Vit+1/2at squared

The Attempt at a Solution

I've been trying to figure out Fa since i know the coeffiecient. But since i cannot find time I'm trying to use another equation to find distance. I'm really bad at finding these things without a mass given becaue I'm so used to that. Thanks for your help!

 

[hage567]

What equation are you using to find the frictional force? That is the only force acting horizontally on the box. So it is the the net force in F_{net} = ma. Use this approach to find the deceleration of the box. Once you have that, you can apply kinematics to figure out how far the box moved.

You don't need to know the mass, it will cancel out in your equations.

 

[Senjai]

i don't understand how mass can cancel out, howe can you use Fnet=ma with two unknowns?

 

[hage567]

i don't understand how mass can cancel out, howe can you use Fnet=ma with two unknowns?

What do you think is the other unknown?

 

[Hells_Kitchen]

keep in mind that after the moment the box is given the push the only force acting on the object in the x- direction is friction from there you can apply

SumF = ma
-f = ma
-(mu)mg = m*a
therefore a = - (mu) g

Now as hage567 said use kinematics to find the displacement of the box.

 

[Senjai]

What do you think is the other unknown?

i don't see how you use F=ma to solve this question, you don't have any of the variables. although i don't see how to solve it myself..

 

[mortho]

well my teacher really hasn't talked much about kinematics or maybe she has..we don't ever use the book, just her notes. But you i used the acceleration which i got -1.76 m/s2 and used it to find time t=-2Vi/a and got 4.77 s and plugged the numbers into x=Vit+1/2at2 and got 40.0 m is that correct?

 

[hage567]

i don't see how you use F=ma to solve this question, you don't have any of the variables. although i don't see how to solve it myself..

After the initial push the only force acting on the block is the frictional force, f = \mu_k N, where N is the normal force (in this case it is equal to the weight of the block, mg).

So put that together with F = ma, the mass cancels and there is only one unknown left, which is a.

 

[mortho]

is 40.0 m correct> ?

 

[Senjai]

After the initial push the only force acting on the block is the frictional force, f = \mu_k N, where N is the normal force (in this case it is equal to the weight of the block, mg).

So put that together with F = ma, the mass cancels and there is only one unknown left, which is a.

oops, I am tired.. thanks lol so so ma = uN

N = mg

ma/mg mass cancles out, a/g = u

so a = g(u) ?

 

[hage567]

well my teacher really hasn't talked much about kinematics or maybe she has..we don't ever use the book, just her notes. But you i used the acceleration which i got -1.76 m/s2 and used it to find time t=-2Vi/a and got 4.77 s and plugged the numbers into x=Vit+1/2at2 and got 40.0 m is that correct?

t=-2Vi/a

Where is that 2 coming from? Find the right time, and your second equation should give the correct result.

But:

You can do this without finding the time. Look at your kinematic equations and see which one has what you know, and the one thing you need. You know the initial and final velocities, right?

 

[Senjai]

oops, I am tired.. thanks lol so so ma = uN

N = mg

ma/mg mass cancles out, a/g = u

so a = g(u) ?

sorry for another interruption, so after a = gu

solve for delta t using a = delta v/delta t
then solve for delta x using v = delta x/delta t ?

 

[hage567]

sorry for another interruption, so after a = gu

solve for delta t using a = delta v/delta t
then solve for delta x using v = delta x/delta t ?

Yes, you can find t using your first equation. But to find x, you must use an equation that involves acceleration, since the block is not moving at a constant velocity.

But, like I said already, you don't actually need to solve for the time directly. There is a kinematic equation you can use that will let you solve for x without knowing what t is. Perhaps it is good practice for you to try it both ways!

 

[mortho]

Ok ..i don't have Vf though..if i use (Vf2-Vi2)/2a and to find Vf i need to have x don't i? hm..let me look some more or think about what i can use to find Vf ..so far i could only think of squareroot Vi+2ax ..

 

[Senjai]

Oh! i see! you must use the formula: \Delta X = V_{1}\Delta t + \frac{1}{2}a(\Delta t)^2 ??

 

[hage567]

Ok ..i don't have Vf though..if i use (Vf2-Vi2)/2a and to find Vf i need to have x don't i? hm..let me look some more or think about what i can use to find Vf ..so far i could only think of squareroot Vi+2ax ..

The block comes to rest, so Vf = 0.

 

[hage567]

Oh! i see! you must use the formula: \Delta X = V_{1}\Delta t + \frac{1}{2}a(\Delta t)^2 ??

Yes, you can use this equation if you are going to find the time first.

 

[Senjai]

Awesome, Thanks a bunch!

 

[mortho]

how do i find the right time?

 

[mortho]

oh ..sorry i didn't see the second page okies..so it's 5.01 m?

 

[hage567]

how do i find the right time?

You can use a_{ave} = \frac{\Delta v}{\Delta t}

 

[hage567]

oh ..sorry i didn't see the second page okies..so it's 5.01 m?

Looks good.

 

[mortho]

Thanks Again!