Calculating the final speed w/ work & friction

[Caleb S]

The problem is asking me to find the final speed of a 1100 kg car traveling at 24 m/s through 18m of mud, where the resistive force on the car is 17000 N.

I don't actually know how to go about doing this, so any pointers in the right direction would be super helpful.

 

[haruspex]

The problem is asking me to find the final speed of a 1100 kg car traveling at 24 m/s through 18m of mud, where the resistive force on the car is 17000 N.

I don't actually know how to go about doing this, so any pointers in the right direction would be super helpful.

Please use the template and include any standard equations you may have been taught that appear relevant.

 

[scottdave]

Do you know about Conservation of Energy? The car has a Kinetic Energy. The force of the mud does some work. Then the car has a different Kinetic Energy.
http://hyperphysics.phy-astr.gsu.edu/hbase/work.html#wepr

 

[Caleb S]

Please use the template and include any standard equations you may have been taught that appear relevant.

There's no template included.

Equations that I think might be helpful:
Work = m*a*x
Kinetic Energy = .5*m*v^2
Kinematics equations

 

[haruspex]

There's no template included.

See item 2 at https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/

Kinematics equations

You mean kinetics. Kinematics is something else.
Yes, kinetics equations can be used here but your other suggestions are better:

Work = m*a*x
Kinetic Energy = .5*m*v^2

Right. So what can you say about the change in KE in passing through the mud?

 

[Caleb S]

(There is no template for the problem. It is given to me in a text form.)

The kinetic energy decreases as the vehicle passes through the mud due to the lowering of the velocity...

I know how to calculate the kinetic energy of the vehicle before it enters the mud. I just don't see how the resistive force of friction relates to the final velocity.

 

[haruspex]

There is no template for the problem. It is given to me in a text form.

There is a standard template provided by this forum for posting homework questions. My understanding is that it appears automatically when you create a thread, which implies you deleted it.

I just don't see how the resistive force of friction relates to the final velocity.

One of the equations you posted in post #4 expresses work done in terms of a force and a distance.

 

[Caleb S]

Oh wait I just realized that there is an equation for the work done by friction: W = -F*d

 

[haruspex]

Oh wait I just realized that there is an equation for the work done by friction: W = -F*d

Right - that is an example of the m.a.x equation you posted earlier, since F=m.a.
So can you connect that with the KE?

 

[Caleb S]

I think so... it would be something along the lines of KE - W_f = KE_net, correct?

 

[Caleb S]

I think so... it would be something along the lines of KE - W_f = KE_net, correct?

And from there you could calculate the final velocity by running it back through the kinetic energy formula, right?

 

[Mark44]

You mean kinetics. Kinematics is something else.

Kinematics is correct. See this wikipedia article: https://en.wikipedia.org/wiki/Kinematics
From the linked article:

Kinematics is a branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that caused the motion.

 

[haruspex]

Kinematics is correct. See this wikipedia article: https://en.wikipedia.org/wiki/Kinematics
From the linked article:

That's why this problem is kinetics (it concerns masses and forces), not kinematics.
As that article goes on to explain, kinematics can be thought of as the geometry of motion. E.g. if you have a linkage or system of pulleys, it is the system of equations that relate the relative positions of its parts as the system changes shape (and consequently, relates their velocities and accelerations).

 

[haruspex]

I think so... it would be something along the lines of KE - W_f = KE_net, correct?

Yes.

 

[Mark44]

That's why this problem is kinetics (it concerns masses and forces), not kinematics.

I stand corrected.