Question

‘A model truck of mass 1.20kg rolls down a ramp angled 30° to horizontal from rest at a height of 0.8m into a tray of sand. Assuming no loss of friction before hitting the sand, what work is done on the truck by the sand and what is the average force of friction the truck experiences in the sand.’ What would be the best way to solve this? Thanks

Answer 1

Hi! I'll give you a hint to start with, and we can work from there!

For the work done, you can consider mechanical work to be the change in the kinetic energy. That is the "work - kinetic energy theorem." It is one of many tools to find the work done, and it's the right tool for the job in this problem. How can you find the change in kinetic energy, though? What is the kinetic energy before versus after?

Answer 2

Hi,
That is where I am getting stuck at the moment. The formula for kinetic energy is \[\frac{ 1 }{ 2 }mv ^{2}\] However with the data I am given I do not have my velocity, nor as far as I can see do I have the means to find said velocity, in order to find Ek and then my Work Done. Thanks

Answer 3

Right! So, \(E_k=\frac12mv^2\) is one way to find kinetic energy!

But, if you assume that mechanical energy is conserved, then you can look at the change in potential energy! Since we assume no loss of friction on the incline, it looks like they're hinting that it's okay to use conservation of mechanical energy. Does that help a little?

Answer 4

So,
I have the required data to use my formula for Potential Energy, \[Ep = mgh\]If I am understanding this right, I can also assume that \[\Delta Ep = Work Done\] Am I right? In which case I would use Initial Ep = (1.2)(9.8)(0.8) and my
Final Ep = (1.2)(9.8)(0?)

Answer 5

Right!

Answer 6

But you'll notice something a little weird with the sign.

Answer 7

When you use math in physics, it always comes from some meaning. And we always have to be careful of that.

What is weird is that the work is the change in kinetic energy. And the change in kinetic energy is opposite the change of potential energy. And so work is opposite the change in potential energy.

Answer 8

I pumped in the data and got an Initial Ep of 9.408 and a Final of 0, meaning my Delta Ep will be -9.408. So therefore my work done will be 9.408?

Answer 9

That's the change in kinetic energy going down the ramp, though. It goes from 0 to whatever at the bottom of the ramp. Then it goes from that "whatever" to 0 when interacting with the sand. So, we want to look at opposite of the kinetic energy going down the hill. That is equal to the change in potential energy, so you were right with your equation.

Answer 10

The work is negative, I believe. I'm sorry if I mislead you by not typing out the whole thing at once.

The question looks for
"what work is done on the truck by the sand"
this is just like
"what work does the sand do on the truck," in case that helps.

Well, the sand takes energy, it doesn't give it. The sand takes energy. That's like giving negative energy. Taking a cookie is like giving negative one cookie, and energy is like cookies, so... Just messing, but that is how math works.
So, we do expect the answer to be negative.

Answer 11

There's actually a lot of basics compiled into this one solution. So, if you have the time and interest, you might like to read this again sometime!

Anyway, you have the answer concerning the work, now!

Now, How to get the average force?

Answer 12

So I am right in thinking the work done on the truck is -9.408?

Answer 13

Yep! :)

Answer 14

At least, I think you're right :)

Answer 15

Great :)
Now as for the Av force, I am not sure about where to start with that one..

Answer 16

Okay! It's another trick! Another physics "tool" that you want to remember to work on problems!

"work" on problems.. I almost made a pun.

I'm not sure about this one.
If we how far it goes into the sand, the trick is that work is the force times the infinitesimal change in distance.
If you don't know the calculus, that's fine. If you do, then \(W=\int F\dot\ \text dx\).

Either way, the non-infinitesimal version is \(W=\overline F\dot\ \Delta x\)

But we need to know how far the truck goes into the sand for that...

Answer 17

Unfortunately the question doesn't give anything aside from the weight, initial height and angle. Nice pun

Answer 18

I don't know, sorry!

We found the work done by the sand on the truck if it stops the truck after rolling down that hill...

But we need to know how long of a distance the sand enacted a force on the truck to find the average force...

Answer 19

hm, no problem. While you are here, I have two simple theory questions regarding newtons laws that I just want to double check
a) A person in a lift feels lighter as it rapidly decelerates to rest when approaching the top floor. Which of Newtons laws of motion would be used in this scenario and what are the physics behind it?
b) A heavy arrow leaves a bow at a slower velocity then a lighter arrow when using the same string tension. Same question

Thanks

Answer 20

Okay! All laws apply to all examples, generally, because they're always in effect!

What do you have?

Answer 21

All laws always apply, but they are not always useful.

Answer 22

I simply need the most prominent law used. I don't have my own answers yet, just wanted to get your opinion. There is a third one which I have answered:
When walking at a constant speed you change direction to the left

For that I answered First Law, reason being that every object of uniform motion will remain that way until an external force is applied to it. If whilst walking you decide to turn left then you apply a force against the ground (and it pushes back) you will be affected.

Answer 23

Ah.. Just one prominent law.. I don't know how to choose! So, you change direction from a constant speed. I would agree that the first law is definitely insightful, and it suggests that an external force was applied. And I see you mentioned the third law in parenthesis, because it describes the force that changes your momentum. So, this is like first and third, but I guess we can go with first, the problem focuses on changing direction more than walking, I guess..

For (a), I will look at the key to finding which law is prominent to be "feels lighter."
When "feeling," we register the pressure, which is related to the force. So, the force...
I like all of the laws... this is silly....
Anyway, the force we feel is the force we exert. Since the lift is accelerating downward... this is tough for me to describe at the moment...

Answer 24

I believe we need to use the other laws to get to the third law, because of the feeling.

For (b), it talks about heavy and light. So we are concerned with mass. Only one law deals with mass, so that's it. Hint: the tension relates to the force, and average tension over the movement will be the same.

Answer 25

Due to the context of other questions within the task I am doing at the moment, I would say the lift question is referring to apparent weight

Answer 26

For (a), since the lift is accelerating downward, it is not supporting the total weight of the person. Thus, the normal force is less. By the third law, the person feels a lesser force than normal, and considers himself or herself to feel lighter.

Answer 27

If you wanted to mention apparent weight,
since the apparent weight of the person is less in this situation, it is implied that the normal force is less and so the person feels perceives a lesser downward force by the third law.

You can mix and match :P

Answer 28

I'm off to bed. Take care!

Answer 29

Thank you so much, I wouldn't have been able to complete what I have done without your help. Take care

Answer 30

You're welcome! If you have questions, please let me know! It's a lot easier to learn this stuff one on one, from my experience.

Sorry if I was unclear at all earlier. I was a little tired!