Finding Velocity and Distance of a raised mass

[tjb5492]

Okay so I am just basically double checking my work.

Problem. A 2kg block is held from a massles rope of height .3m , and is aloud to swing as a simple pendulum. Once it reaches the bottom of the the swing it is released. coming to a rest of displacement d. the kinectic friction uk= .6
using conservation of energy what is speed v when it reaches the table top?
b) find D

What I did. I calculated work from gravity Wg= F D => (2)(9.8) (.3) = 5.88 J

I used an equation w= 1/2 mv^2
5.88= 1/2 (2)(v)^2
v^2=5.88
v= 2.42 m/s

Solving for Part B

F=ma
5.88=(2)(a)
a= -2.94

I then used Uk and fn to get friction force ... .6(9.8)(2) = 11.76 n

I used my chapter 2 equations to then find D

V^2= vo^2 + 2A (xf-xi)
0= 2.42^2 + 2(A) (x)
x= .99

My doubts...
I am confused on how to find acceleration. I know that F=ma

but wouldn't 5.88 be a force because its finding how much work is being done, assuming there is no air drag it shows how much force is being excerted at the very bottom.??

 

[gneill]

How long is the string? How high is the mass raised before starting to swing (or equivalently, what is the angle of displacement from vertical of the string)? How high above the table is the bottom of the pendulum arc?

Or is the bottom of the pendulum arc at the table level? I suppose that would make sense given that you seem to be dealing with friction...

 

[tjb5492]

the distance the object is raised is .3 meters... length of string is not given...
basically we have a block .3 meters above a table top. its in pendulum form. once it hits the table it is released. we need to find speed. and also find how far it goes. the given info.
mass 2kg block
uk= .6
frictionless rope..

It says that we need to solve using conservation of energy . i calculated PE by mgh.
but after that i didnt know how to find accel. i kno that at the bottom where it is released the potential energy is 0 and KE is = PE. but i need to find speed of block when it is released and slides acros table, and also find how far it goes using Uk..ah.

 

[gneill]

the distance the object is raised is .3 meters... length of string is not given...
basically we have a block .3 meters above a table top. its in pendulum form. once it hits the table it is released. we need to find speed. and also find how far it goes. the given info.
mass 2kg block
uk= .6
frictionless rope..

It says that we need to solve using conservation of energy . i calculated PE by mgh.
but after that i didnt know how to find accel. i kno that at the bottom where it is released the potential energy is 0 and KE is = PE. but i need to find speed of block when it is released and slides acros table, and also find how far it goes using Uk..ah.

You can bypass all the velocity and acceleration stuff by sticking to conservation of energy.

How much work is done by friction in order to bring the block to a halt? Can you calculate the force due to friction? What's an expression for the work done by a force over a given distance?

 

[tjb5492]

Work done by friction to make object stop would have to be vf equal zero. I'm very lost on how to do this we just learned this and have a test tomorrow

 

[gneill]

Work done by friction to make object stop would have to be vf equal zero. I'm very lost on how to do this we just learned this and have a test tomorrow

Work is energy. How much energy does the block start out with when it first reaches the table? What form is that energy in?

 

[tjb5492]

okay so the work done is that by gravity. work by gravity is (9.8)(2) (.3)= 5.88 J

I would guess it is Kinetic Energy at this point?

 

[gneill]

okay so the work done is that by gravity. work by gravity is (9.8)(2) (.3)= 5.88 J

I would guess it is Kinetic Energy at this point?

Yes. The block begins its slide with that amount of kinetic energy. That energy is going to be turned into heat energy by friction. When it's all converted, the block stops moving.

What's an expression for the force due to friction?

 

[tjb5492]

fn(uk).. so it would be (2)(9.8) ( .6) = 11.76 N
so with this i tried to do 5.88-11.76=ma
a=-2.94 which would make sense because it is slowing down.
BUT i second guessed myself because f=ma.. and i know 5.88 is the amount of energy, is that equivalent to the force acting in that instant?

 

[gneill]

fn(uk).. so it would be (2)(9.8) ( .6) = 11.76 N
so with this i tried to do 5.88-11.76=ma
a=-2.94 which would make sense because it is slowing down.

No, you're mixing Joules and Newtons. One is an energy and the other is a force.

BUT i second guessed myself because f=ma.. and i know 5.88 is the amount of energy, is that equivalent to the force acting in that instant?

Forget acceleration. You're working with energy.

You've calculated the frictional force that will act as long as the block is moving (11.76N).
Now, what is the expression for work done by a given force (you've already used it to find the work done by gravity!).

 

[tjb5492]

could it be f (dot) d

but this is where i am confused at. i know its fd cos theta. but what is theta? is theta 0 or 180? or do we not worry about theta?

 

[SHISHKABOB]

well, how is the direction of the force of friction on an object oriented relative to the direction in which an object is traveling?

 

[tjb5492]

180 degres because friction is opposing the direction its moving. so cos 180= -1

so we are left with -11.76d
because we are not given do. Now its almost time to solve for D.

but what exactly do we do to solve for D? Fw=-11.76d
where Fw = work done by friction

we now have Fw=-11.76d and Ke= 5.88

 

[gneill]

could it be f (dot) d

but this is where i am confused at. i know its fd cos theta. but what is theta? is theta 0 or 180? or do we not worry about theta?

Friction force is always in the opposite direction to velocity (motion). So the angle would be 180 degrees. But really, all you have to deal with here is the magnitude, f*d, just as you did to find the work done by gravity.

So write your equation for the magnitude of the work done, setting it equal to the available energy (Kinetic Energy).

 

[tjb5492]

W= Delta k= 1/2mv^2-1/2mv0^1/2
available energy is 5.88

honestly i don't know how to set up this equation guys. I've really been trying to understand, and i pretty much have up until this point. I know that kinetic energy is 5.88 J but what am i settin this equal to? i know you said magnitude of work done.

 

[SHISHKABOB]

here's a simple theorem that might help: The work-energy theorem says that the work on the object is equal to the change in kinetic energy of the object.

so what is doing work on the object as it travels across the surface of the table?

 

[gneill]

W= Delta k= 1/2mv^2-1/2mv0^1/2
available energy is 5.88

honestly i don't know how to set up this equation guys. I've really been trying to understand, and i pretty much have up until this point. I know that kinetic energy is 5.88 J but what am i settin this equal to? i know you said magnitude of work done.

W = F*D

Work is energy. Energy available is 5.88J . Force doing the work is 11.76N.

5.88J = 11.76N*D

D= ?

 

[tjb5492]

d= .5

 

[gneill]

d= .5

Ta-da!

Note that you could have skipped a lot of the intermediate converting things here and there. If you know your source of energy and how it's eventually "spent", you can go right from start to finish and skip the racecourse, so to speak. It's one of the powerful features of conservation theories.

In this case you have energy supplied by gravity, ending up as work done by friction. So:

m*g*h = (μ_k*m*g)*d

Cancelling where required,

h = μ_k*d

d = h/μ_k