How Do I Calculate the Final Velocity of a Trolley?

[hutchiee]

Hi, I've been given an assignment from college which has me totally baffled! (Ashamed to say this is my weakest subject ha)
I have read the rules and I do not wish for my work simply done for me, just a explanation would be brilliant! Or a similar question would be great so I could practice it myself!

Question:

A trolley of mass 2.5kg is pulled along a flat bench by a force of 5N. When the trolley has traveled 4m find:
A How much work has been done.

Am i right in thinking WORKDONE(Nm)= FORCE(N)XDIST.MOVED(m)

So would that simply be 5Nx4m?

B How much kinetic energy it has gained

From my notes I have:

K.E.= mv^2
------
2 where m=mass(kg) & v=velocity(m/s)

If that is so then how do I calculate how much it has gained?

C It's final velocity

Would KE=mv^2 be correct in working this out? Also would the KE be representive of how much it has gained, or overall?

I know this may seem simple to some of you, but this really is out of my comfort zone. Any help would be greatly appreciated, thanks!

 

[Doc Al]

A How much work has been done.

Am i right in thinking WORKDONE(Nm)= FORCE(N)XDIST.MOVED(m)

So would that simply be 5Nx4m?

Yes, simple as that. (Work and energy both have units of Joules.)

B How much kinetic energy it has gained

From my notes I have:

K.E.= mv^2
------
2 where m=mass(kg) & v=velocity(m/s)

If that is so then how do I calculate how much it has gained?

Yes, assuming that the applied force is the only force doing any work on the object. (No friction, for example.)

C It's final velocity

Would KE=mv^2 be correct in working this out? Also would the KE be representive of how much it has gained, or overall?

Once you know the final KE, you can figure out the velocity. (Don't forget the ½! KE = ½mv².)

Looks like you're doing pretty well.

 

[hutchiee]

Thanks for the reply, still have not managed to complete this yet ha.

If KE=mv^2, how do i know the velocity(m/s)?

Sorry again, this is nothing like anything I do day to day so it's like a foreign language to me!

 

[gneill]

With no other forces operating (such as friction or gravity), the work done by the applied force will equal the change in kinetic energy: ΔKE = ½mv² .

If the trolly starts from rest (initial velocity is zero), then the change in kinetic energy will be equal to the final kinetic energy of the trolly.

 

[rwisz]

Hello, it seems like you have a good understanding of the formulas and concepts involved in this problem. To answer your questions:

A) Yes, you are correct in thinking that work done (in joules) is equal to force (in newtons) multiplied by distance moved (in meters). So in this case, the work done would be 5N x 4m = 20 joules.

B) To calculate the kinetic energy gained (in joules), you can use the formula KE = 1/2 x mass (in kilograms) x velocity^2 (in meters per second). So in this case, it would be KE = 1/2 x 2.5kg x v^2. Since we do not know the final velocity yet, we cannot calculate the exact amount of kinetic energy gained. However, we can use the work-energy theorem which states that the work done on an object is equal to the change in its kinetic energy. So in this case, the kinetic energy gained would be equal to the work done, which we calculated in part A.

C) To find the final velocity, we can rearrange the kinetic energy formula to solve for v. So it would be v = √(2 x KE/m). Plugging in the values we have, it would be v = √(2 x 20 joules/2.5kg) = √(16) = 4 m/s. So the final velocity of the trolley would be 4 meters per second.

I hope this helps and good luck with your assignment! Remember, practice makes perfect so keep practicing and asking for help when needed.