Coursework in tomorrow, Completely Stuck

[simonegdell]

Homework Statement

a body mass of 5kg starts off from rest and is acted upon a force of 5N over a distance of 2m. if there is no frictional forces make use of the principle of the conservation of energy to determinei ts final speed

Homework Equations

above is all i have.

The Attempt at a Solution

didnt know where to start? any help?

 

[turdferguson]

Use the work-energy theorem

 

[simonegdell]

sorry, new to the whole engineering thing, my first week today, could someone give me a lot more help with this sorry.

 

[dontdisturbmycircles]

Remember that $$F=ma$$. Where force is F, m = mass, a = acceleration.Thus you can find out the acceleration.

Then you need to find out which kinematics (The study of motion in one direction) equation you need to use. Because your assignment is due tommorow, I'll pick it for you.

Try using $$v_{f}^{2}=v_{i}^{2} + 2ad$$

Where $$v_{f}$$=final velocity , $$v_{i}$$=initial velocity, a = acceleration, and d = distance.

Here are the other three kinematics equations that you will need to memorize, just in case you haven't done so already.

$$d=v_{i}t+\frac{1}{2}at^{2}$$
$$v_{f}=v_{i}+at$$
$$d=\frac{v_{i}+v_{f}}{2}*t$$

memorize them!

Most of these formulas can be understood quite easily by looking at a velocity-time graph. Since acceleration is constant, velocity will be linear. The slope of the velocity time graph is the acceleration, so rise/run = a = $$\frac{v_{f}-v_{i}}{t}$$. Etc. In fact that may be the best way to memorize them.

 

[gabee]

Remember that $$F=ma$$. Where force is F, m = mass, a = acceleration.Thus you can find out the acceleration.

Then you need to find out which kinematics (The study of motion in one direction) equation you need to use. Because your assignment is due tommorow, I'll pick it for you.

Try using $$v_{f}^{2}=v_{i}^{2} + 2ad$$

Where $$v_{f}$$=final velocity , $$v_{i}$$=initial velocity, a = acceleration, and d = distance.

Here are the other three kinematics equations that you will need to memorize, just in case you haven't done so already.

$$d=v_{i}t+\frac{1}{2}at^{2}$$
$$v_{f}=v_{i}+at$$
$$d=\frac{v_{i}+v_{f}}{2}*t$$

memorize them!

That's one way to do it, but his question calls for it to be done with conservation of energy equations. Remember that work done by a force is given by multiplying the magnitude of the force by the distance the object moves in the direction of the force:

$$W = Fs$$

And like turdferguson said, use the work-kinetic energy theorem, which states that the amount of work done is the change in kinetic energy of the system

$$W = \Delta K$$

and kinetic energy of an object is given by

$$K = \frac{1}{2}m v^2$$

So, work is done by the 5 N force to increase the kinetic energy of the object. You can find how much work was done, and therefore how much the kinetic energy increased--and therefore how much the velocity increased.

 

[dontdisturbmycircles]

Ahhhh true enough, thanks gabee. I should probably stick to learning the stuff before I start to help with homework problems. :-)

 

[simonegdell]

i know how to get kinetic energy but don't know how use that to find its final speed. i have 3 hours to go! ahhhh!

 

[Hootenanny]

How do you define the work done by a [constant] force?

 

[simonegdell]

am i trying to work out work done? WD = FD, WD = 5N x 2m WD = 10J?

 

[Hootenanny]

am i trying to work out work done? WD = FD, WD = 5N x 2m WD = 10J?

Correct! So where does all this work done (energy) go to?

 

[simonegdell]

into moving the 5kg mass (accelerating from 0?)

 

[Hootenanny]

into moving the 5kg mass (accelerating from 0?)

Yes, good! So the work done is converted into ________ energy.

 

[simonegdell]

kinetic energy? what formula do i use? 1/2 mv^2 is what i have, but i don't have v^2,

 

[Brewer]

Thats the point. You need to work out v^2.

 

[simonegdell]

is that v^2 = 1/2 mass divided by Kintetic energy

 

[Hootenanny]

Okay, you know that the work done by the force is 10J and that this is converted in the kinetic energy of the body. Therefore;

$$10 = \frac{1}{2}mv^2\hspace{1cm}\text{Find }v$$

Do you follow?

 

[simonegdell]

i not sure how to rearrange this but is it somehting along the lines of sqrt of 1/2ke / m

 

[Hootenanny]

i not sure how to rearrange this but is it somehting along the lines of sqrt of 1/2ke / m

Close;

$$v = \sqrt{\frac{2E_k}{m}}$$

 

[simonegdell]

so its final velocity is 2m/s?

 

[Hootenanny]

so its final velocity is 2m/s?

Correct! text too short

 

[simonegdell]

thankyou very much! you have been extremely helpful!

 

[BobG]

Edit: Oops. There was a second page. I guess you already figured it out.