Question about distance using Newton's Laws of Motion

[sona1177]

Homework Statement

A crate pushed along the floor with velocity Vi slides a distance d after the pushing force is removed.

a)If the mass of the crate is doubled but the initial velocity is not changed, what distance does the crate slide before stopping? Explain.

b) If the initial velocity of the crate is doubled to 2vi but the mass is not changd, what distance does the crate slide before stopping? Explain.

Homework Equations

F=ma
F=max (object is moving horizontally)

The Attempt at a Solution

a) Fnetx=nx + wx + tx where t stands for the pushing force, n is the normal force, and w is weight pulling down on the object.
nx=0
wx=0
Tx=max
Tx/m=ax

Kinetics Equation:
V2=Vi2 + 2a(Xf-Xi)
0=Vi2 + 2a(Xf-Xi)
-Vi2=2(Tx/m)d (where d stands for Xf- Xi)
Doing algebra: d=-mVi2/ (2*Tx) (in this equation only the Vi is squared, in the denominator, the answer is (2 times Tx). I hope that is clear.

b) -m * 2 * Vi2/2 *Tx (in this equation its 2 times the initial velocity squared divided by 2 times Tx) therefore the 2's cancel out and I get the same answer as I did for part a -m* Vi2/Tx=d

I hope this makes sense. Is my work and answer correct? I'm not sure if this was the answer the book was looking for. Thank you kindly for taking the time to help me.

 

[kuruman]

First off, all the pushing force does is to give the crate an initial velocity and then it is removed. So Tx =0 and should not be a part of the subsequent motion of the object.
Secondly, you have omitted the force of kinetic friction. How does that figure into the picture?

 

[sona1177]

First off, all the pushing force does is to give the crate an initial velocity and then it is removed. So Tx =0 and should not be a part of the subsequent motion of the object.
Secondly, you have omitted the force of kinetic friction. How does that figure into the picture?

The crate slides a distance d only after the pushing force is removed. So doesn't the pushing factor have to be included? And my book says to ignore friction unless the problem asks to include it. This problem does not. And if I don't include the pushing force, the answer becomes 0/m=ax, meaning acceleration is zero since the normal force and weight force do not have x components. Can you please help guide me with this? Thank you kindly.

 

[sona1177]

Please Help! Thank you.

 

[dulrich]

kuruman is right on both counts.

You should ignore the pushing force...in a sense it's built into the initial velocity. Also, you have to assume there is some kinetic friction because parts (a) and (b) imply the object comes to a stop.

 

[kuruman]

The crate slides a distance d only after the pushing force is removed. So doesn't the pushing factor have to be included?

No. Only the velocity that the crate has at the instant the force is removed needs to be included.

And my book says to ignore friction unless the problem asks to include it. This problem does not.

It does implicitly. What stops the crate if not friction?

And if I don't include the pushing force, the answer becomes 0/m=ax, meaning acceleration is zero since the normal force and weight force do not have x components. Can you please help guide me with this? Thank you kindly.

Why 0 = ma_x? How about friction = ma_x?