Kinetic and Gravitational Energy Question

[NeomiXD]

A dim but resourceful student has a summer job in a factory moving boxes. The 50 kg student finds that if he runs at a speed of 5.8 m/s and then dives head first into the boxes, he can move them across the floor. If the boxes have a mass of 65 kg and the force of friction between the boxes and the floor is 382 N, how far will he move the boxes after hitting it six times?

How am I suppose to solve this question?

m of student = 50 kg
m of boxes = 65 kg
v = 5.8 m/s
force of friction = 382 N

 

[Pi-Bond]

How do you relate force with work? Can you do this to find the work done by the frictional force?

 

[NeomiXD]

Work is done whenever a force causes an object to move...but I don't understand how the force of friction is involved in the process of finding the distance for this question (what formulas are used?).

 

[Pi-Bond]

What's the formula for work?

 

[NeomiXD]

w = F x d

 

[Pi-Bond]

You have the force (F), and you need to find the distance (d). Do you know where w could come from?

 

[NeomiXD]

Gravitational potential energy?

 

[Pi-Bond]

But you don't have any information on height - you have information on the student's velocity. This should tell you the type of energy you need to use for w.

 

[SteamKing]

I think the floor might be level, so the gravitational potential energy is probably not changed when the boxes are moved.

Say, what about that fellow who runs and dives head first into the boxes? Do you think there might be something wrong with him?

 

[NeomiXD]

But you don't have any information on height - you have information on the student's velocity. This should tell you the type of energy you need to use for w.

Kinetic energy? So you have to find the Kinetic energy of the student? If so then the Kinetic energy would be 145 J.

 

[Pi-Bond]

Yes, you can equate it with the work done by the frictional force to find the distance (d). This works because the kinetic energy is dissipated by the frictional force.

 

[NeomiXD]

So I should now use the formula d = W/F, where F is the force of friction (382 N)?

 

[gneill]

A point for the helpers in this thread to consider: When the student dives headlong into the boxes, will this be considered an elastic or an inelastic collision? Will it make a difference to the outcome? If so, how?

 

[Pi-Bond]

I assume an elastic collision so the kinetic energy is constant in the collision. Of course, this won't be possible in real life, but the question doesn't mention anything either!

@NeomiXD - Yes the d you get should be for one dive. You can multiply your answer by 6 to get the total distance.

 

[gneill]

I assume an elastic collision so the kinetic energy is constant in the collision. Of course, this won't be possible in real life, but the question doesn't mention anything either!

In an elastic collision the KE may be conserved but it will be split between the box and the student...

 

[NeomiXD]

I assume an elastic collision so the kinetic energy is constant in the collision. Of course, this won't be possible in real life, but the question doesn't mention anything either!

@NeomiXD - Yes the d you get should be for one dive. You can multiply your answer by 6 to get the total distance.

I was just wondering about the mass of the boxes (65kg). Doesn't that affect the distance too?

 

[Pi-Bond]

Yes, but then again the question doesn't say anything about the velocity of the student afterwards. I'd assume the student comes to rest so that all kinetic energy is with the boxes (hence covering maximum distance). I approach these questions with such "ideal circumstances" in mind as discussed, because most of the time that is what is intended.

 

[Pi-Bond]

I was just wondering about the mass of the boxes (65kg). Doesn't that affect the distance too?

The kinetic energy of the student is completely transferred to the boxes, so they have energy equal to $\frac{1}{2} m v^{2}$, where m is the mass of the student, and v is the velocity.

 

[gneill]

Yes, but then again the question doesn't say anything about the velocity of the student afterwards. I'd assume the student comes to rest so that all kinetic energy is with the boxes (hence covering maximum distance). I approach these questions with such "ideal circumstances" in mind as discussed, because most of the time that is what is intended.

The student would only come to rest if his mass were equal to that of the box.

Simplifying assumptions are only useful if they they don't break physical laws . To quote Albert E., "Everything should be made as simple as possible, but no simpler".

 

[Pi-Bond]

True - I didn't consider this. So the correct way to proceed would be to first find the new velocity of the box from the collision, right?

 

[Pi-Bond]

I was just wondering about the mass of the boxes (65kg). Doesn't that affect the distance too?

It seems I made an error in the first part by assuming that the kinetic energy of the student is transferred to the boxes completely. Sorry about that; assuming an elastic collision, you will need to equate the momentum and kinetic energy of the boxes and students before and after the collision to get the kinetic energy of the box. You can substitute that in w in the equation you mentioned. Sorry again!

 

[gneill]

So, to summarize the plot so far...

The problem can be broken down into two parts. The first part involves the elastic collision of the dim student with a box. Paraphrased, this would be, "a 50 kg object moving with a speed of 5.8 m/s collides elastically with a stationary object of mass 65 kg. What is the speed, v, of the second object immediately following the collision?"

The second part is: "A 65 kg object (box) has an initial speed v, sliding horizontally across a surface. If the frictional force between the box and the surface is 382N, how far will the box slide?"

 

[NeomiXD]

So, to summarize the plot so far...

The problem can be broken down into two parts. The first part involves the elastic collision of the dim student with a box. Paraphrased, this would be, "a 50 kg object moving with a speed of 5.8 m/s collides elastically with a stationary object of mass 65 kg. What is the speed, v, of the second object immediately following the collision?"

The second part is: "A 65 kg object has an initial speed v, sliding horizontally across a surface. If the frictional force between the box and the surface is 382N, how far will the box slide?"

So, what formulas are you suppose to use to find the speed, v, of the second object immediately following the collision?

 

[gneill]

So, what formulas are you suppose to use to find the speed, v, of the second object immediately following the collision?

Conservation of momentum, conservation of kinetic energy. Your text or class notes should explain collisions and the use of these conservation laws.