Need help on finding the work done on an Object by the applied force.

[xuhhad]

"A 50 kg crate is being dragged across a floor by a force of 225 N at an angle of 40° from the horizontal. The crate is dragged a distance of 5.0 m and the frictional force is 60 N."

Now from that information i am supposed to find the work done by the applied force on the object.

This is what i have so far. Since W= (F cos θ)Δd, i did W= (225 N cos 40°)5.0 m ≈ 862 N.m
Am i correct

PS this is my first time on physics forum. :)

 

[tiny-tim]

welcome to pf!

hi xuhhad! welcome to pf!

"A 50 kg crate is being dragged across a floor by a force of 225 N at an angle of 40° from the horizontal. The crate is dragged a distance of 5.0 m and the frictional force is 60 N."

Now from that information i am supposed to find the work done by the applied force on the object.

This is what i have so far. Since W= (F cos θ)Δd, i did W= (225 N cos 40°)5.0 m ≈ 862 N.m
Am i correct

yes

(what is worrying you about that?)

 

[Simon Bridge]

yep - hi xuhhad! welcome to PF!
It can be a bit daunting at first - especially when an answer seems to come too easily. As you learn more you'll be able to develop strategies to figure out if you are right or not without having to ask anyone.

I suppose it next asks for the work done by the frictional force?

 

[xuhhad]

Yes it does :/

and again i am stuck sorry for the inconvenience. I really need help.

 

[Simon Bridge]

Funny that :) You have to think it through - what is the definition of work in terms of force? Can you write it in words?

 

[tiny-tim]

hi xuhhad!

you seem to have a lack of confidence

but your original answer, using the correct definition of work done, was fine, so just do the friction in exactly the same way, and then that'll be fine too!

what do you get? ​

 

[xuhhad]

Funny that :) You have to think it through - what is the definition of work in terms of force? Can you write it in words?

hmm the work formula would be W=mg/Δd

 

[Simon Bridge]

Get used to putting the formulas into words and thinking about what they mean.

Your formula has work equal to the vertical force divided by the displacement ... which would mean that the further you drag the crate, the less energy is taken by friction ... does that sound right?

Aren't you also lifting the crate a bit as you pull it? Does that matter?

(Since you have not been given dimensions for the crate, I expect that you are to treat all forces through the center of mass?)

 

[tiny-tim]

hmm the work formula would be W=mg/Δd

no, work done = force "dot" displacement …

(so it can be positive or negative)

learn this definition! ​

 

[xuhhad]

i still don't understand how i can find the frictional force.

 

[tiny-tim]

i still don't understand how i can find the frictional force.

??

the frictional force is given in the question …

it's 60 N horizontally, "backwards"​

 

[xuhhad]

OHHHH
i get it i just have to use the work formula again, something like this correct

W=(-Ff)cos θ * Δd

Thak u for all the help

 

[dimension10]

Yes it does :/

and again i am stuck sorry for the inconvenience. I really need help.

F_kinetic = μ_k F_normal

Does that help?

 

[dimension10]

This is what i have so far. Since W= (F cos θ)Δd, i did W= (225 N cos 40°)5.0 m ≈ 862 N.m
Am i correct

Yes...

This part of your question sounds like:

The problem
A=1, B=2, AB

Relevant equations
AB=AB

Solution
AB=(1)(2)=2
Am I correct?

P.S. I'm not hijacking this thread. I obviously know what 1 multiplied by 2 is. I'm just saying that the problem I wrote is very similar to the problem given by the OP.

 

[xuhhad]

F_kinetic = μ_k F_normal

Does that help?

Yes thank u for posting i already completed the question thank u for the help though

 

[dimension10]

Yes thank u for posting i already completed the question thank u for the help though

Oh..

 

[christpinus]

if the is a force acting opposite to the direction of motion of the crate ( frictional force= 60N) this implies that the total force pulling the crate is now; 225N - 60N =165N.
the reaction force to the floor will be= 50cos40=38.30N
Therefore, total pulling force= 165N - 38.30N= 126.7N.
Now Work done= 126.7 x 5 = 633.5J

 

[christpinus]

Pls i hope you have seen where your errors are coming from. You forgot about the reaction force onthe crate and also the frictional force opposing the force causing the motion of the crate. Thanks