Catapult Question: Force of Friction & Kinetic Energy

[TyErd]

The strain energy of a catapult extended by 1metre is 20J. When x=1m, A 1kg block is catapulted a distance of 2.5metres from the point of release along a horizontal floor.

a. What is the magnitude of the force of friction between the block and the floor?
b. What is the maximum kintetic energy of the block?

 

[joriarty]

Before anyone can help you, you need to show us your attempt at an answer and tell us what you are having trouble understanding.

We won't do your homework for you. We'll only help you understand how to tackle the problem.

 

[TyErd]

honestly i don't know where to start. I know some equations that may have something to do with it but that is all. I am really stuck on this one.

 

[joriarty]

Have you been going over coefficients of kinetic friction recently in class/lectures?

 

[joriarty]

Note my interpretation of the question is that the block is being catapulted horizontally along the floor and doesn't lift off the ground at any stage... (which seems to go against the conventional picture of what a catapult is)

 

[TyErd]

yes your interpration is correct, it doesn't lift off the ground.

 

[joriarty]

Try considering how the energy of the catapult is transferred to the block and then how it is lost due to friction. Using conservation of energy you can easily figure out how much kinetic energy the block will have lost due to friction. How can you then relate that kinetic energy lost to the distance traveled (which you also know) and the force which opposes the direction of motion of the block?

 

[TyErd]

firstly, how do i figure out how much kinetic energy lost due to friction?

 

[joriarty]

All you need to know is that the total kinetic energy transferred to the block is going to be equal to the amount of elastic potential energy the catapult has lost.

At your level of study you can safely assume that all of this energy is transferred to the block gradually, as the block travels the first 1m and it is still being "pushed" by the catapult. This means that the maximum kinetic energy the block has will be less than the total energy transferred to it by the catapult because during that first 1m it is still losing energy to friction. But that information is not important for the first part of the question.

If this information still does not help you finish the first question, that would indicate you do not understand the fundamental physics behind the problem. In which case you should revise your notes and your textbook or consult your teacher/lecturer - there's no point attempting problems if you do not understand the background theory. We don't have all day to help you and I'm not going to just hand you the answers.

 

[TyErd]

I understand the kinetic energy parts, it is friction that i am having trouble with.

 

[joriarty]

Well think of it this way - the friction force is going to directly oppose the motion of the block. You know the total energy lost to this friction force and you know that this energy is lost over a distance of 2.5m. What formula can you use here?

 

[TyErd]

W=Fx

 

[TyErd]

ok so the force due to friction is 8N. The 2nd part is max kinetic energy, so I am thinking i do total energy - energy lost due to friction

 

[TyErd]

problem is how to calculate energy loss due to friction.

 

[joriarty]

Your answer of 8N sounds right. For the second part - the total energy of 20 J is equal to the total energy lost due to friction (which is 20 J in total as well). If I interpreted you correctly and that was the approach you were considering, that won't work.

I am presuming that you may assume that over the 1m distance where the block is still in contact with the catapult, the force exerted on the block is constant. If this is the case, when is the block going to have maximum kinetic energy? (ie at what distance traveled will it have maximum kinetic energy?) Then consider how much energy it has already lost due to friction over that distance.

I have to go now and won't be able to check back for a day or two so you're probably on your own from now unless someone else helps you - but you're almost there. Good luck.

 

[TyErd]

ok thnx for the help

 

[TyErd]

any1 else want to help me out?