Problem with friction that is mathematically feasible but conceptually confusing.

[guncamp]

In my high school physics class we are working with energy. We had the following problem that puzzled us slightly:

 

A 70 kg box is pulled horizontally with a constant force of 200N. It is pulled for the first 10 meters across a frictionless surface, and then is pulled another 10 meters over a surface with a coefficient of of friction that is equal to .3. Find the final velocity of the box.

 

so the the total energy done is equal to:

 

(200)(10) + (200)(10) - (70)(9.8)(10)(.3) = 1954

 

Therefore, it is equal to the final kinetic energy 1954= 1/2 (70)(v²)

 

v= 7.5 m/s.

 

This is all fine and dandy, but the problem appears to arise in the second part where the box is moved across the surface with coefficient of friction of .3.

 

μ=.3

 

(9.8)(70)(.3) = 205.8 N of frictional force is acting against the 200 N of constant horizontal force applied to the box. Now, since the amount of kinetic friction is greater than the force being applied, how could the box traverse the final 10 meters?

 

The only thing I could think of is the energy built up from the first 10 meters was enough to help carry it across the final 10 meters with friction.

 

Are we missing something here?

 

*Note that I don't have my textbook with me, and I think I provided all the information that was given to me in the problem. Bear with me if I forgot a piece, I did my best to recall all the information.

 

Thanks in advance.

[swansont]

!

Moderator Note

There is a post hidden containing the worked-out solution to the problem. That's not what we do for homework problems.

 

 

 

The only thing I could think of is the energy built up from the first 10 meters was enough to help carry it across the final 10 meters with friction.

 

Are we missing something here?

 

 

Nope. That's it, exactly. The net force will be in the opposite direction of motion, which means it will slow down. Work will be done to reduce the KE.

[studiot]

 

In my high school physics class we are working with energy. We had the following problem that puzzled us slightly:

 

The only thing I could think of is the energy built up from the first 10 meters was enough to help carry it across the final 10 meters with friction.

 

There is another approach, that leads to the same answer, but you said you were considering energy.

 

Since the 200N force is constant it imparts constant acceleration to the box.

 

So if you draw a velocity-time graph;

 

The box starts at time t₀ = 0 at zero velocity v₀=0 and is then accelerated by the constant force with acceleration a₁ for a distance of 10 metres until it enters the friction zone at a velocity of v_1.

The box is then subject to a constant acceleration a₂ (which works out negative) due to the difference between the 200N force and the opposing frictional force for a further distance of 10 metres until it reaches v_2.

 

You have enough information to solve the system this way.

 

Have you covered this method in your studies?

Edited February 13, 2014 by studiot