How Does Friction Affect the Work Needed to Move a Crate Up an Inclined Plane?

[Cole07]

Homework Statement

How much work is needed to push a 112- kg packing crate a distance of 2.6 m up a frictionless inclined plane that makes an angle of 34 o with the horizontal?
Answer: 1595.80
How much work would be required to move the crate the same distance if the coefficient of friction were 0.30?
Answer:?

Homework Equations

W=F*D

The Attempt at a Solution

I don't understand the second part is it saying that with friction what would it be? and if so would you not just subtract 0.30*2.6=0.78 and then add that to the first answer? thanks for any help.

 

[Ja4Coltrane]

okay, first of all for the first part, knowing that work is equal to a change in energy (any type) how much energy do you think you would need to get the block up this ramp?
For the second part, you sort of have the right idea. Knowing the amount of energy needed to get it up without friction, it should make sense that you just have to add to that the work that friction does on the block (or really the opposite of that). Now, you said that W=F*D, so you might be able to go from there?

 

[Cole07]

so would this be right 1595.80+0.78= 1596.580234J ?

 

[cristo]

Cole07; it might be better if you show your working, that is, the equations you used, what values you put in for each term. It's hard to see where you're going if we can't see what you've done. And, of course, simply putting down numbers implies that you expect us to work it out ourselves in order to check your work!

 

[Ja4Coltrane]

yeah, you don't even need to use nubers if you don't want to, just put it in some way so that we can look at your technique. Thanks!

 

[Cole07]

sorry.
to get the 1st answer i drew a diagram of a triangle with the crate on the hypotnuese pointed NE the angle is 34 degrees so i used sin to find the opposite of the angle and then i found the weight of the crate in Newtons which is 112kg*9.8= 1097.6 then i used sin to find the y side of the triangle 1097.6sin(34)=613.7701308N then i used the equation W=F*D
W-613.7701308*2.6 so W=1595.80234J (the first answer) . then for the second question is where I'm stuck but what i think you do is take 0.30*2.6 which is friction force and then add this to the first answer.

 

[Ja4Coltrane]

okay--good job on the first part! (you could have more easily just used W=change in energy=Umax=mgh=mgdsin@ but whatev)
Now, you are not doing the friction correctly. Frictional force is normal force times the coefficient of kinetic friction. So find that and then multiply it by 2.6 to get the work.

 

[Cole07]

i don't understand what you mean to do by this?

 

[Ja4Coltrane]

normal force is the force exerted by the ground on the object (Newtons third law) if an object sits on a flat surface, the force is simply the weight of the object, but because it is inclined, you need to take a component of its weight in the direction perpendicular to the incline.

 

[Cole07]

ok i understand what you are talking about but how do i take a component of its weight don't get that.

 

[Cole07]

do you mean what the hypotnuese is?

 

[Ja4Coltrane]

okay Theta=34 degrees. The weight is mg so try to go from there.

 

[Cole07]

i'm sorry but I'm not getting this my teacher told me we should be able to do all the questions from this assignment, but i just don't get it.

 

[Cole07]

ok i do understand that theta is 34 degrees and that the weight would be 1097.6N.

 

[Ja4Coltrane]

okay, Ill help you out (if it's not too late) the component of weight towards the surface is mgcos34. If you draw some triangles, you should be able to see this. so next, Friction is equal to .3*normal force. with that you have your force so multiply it by the distance to get the work done. does that make sense? By the way--if you are having trouble seeing why I use cosine, imagine that the angle was instead of 34 degrees only 1 degree. you should see that the normal force would be very close to the actual weight and cos1 is extremely close to one (probably like .9999 something). tell me if that helps or not.

 

[Cole07]

would this be what Ja4Coltrane has told me to do?
mgcos34:
(112)(9.8)cos(34)=909.516396
909.516396*.3=272.8549188
272.8549188*2.6=709.4227889J
Q:is the normal force the same as mgcos34?

 

[Cole07]

could someone please tell me if I'm on the right track.

 

[cristo]

Sorry, I didn't read the question properly before replying the first time!

would this be what Ja4Coltrane has told me to do?
mgcos34:
Normal force=(112)(9.8)cos(34)=909.516396
Frictional force=909.516396*.3=272.8549188
Work done to overcome the frictional force=272.8549188*2.6=709.4227889J

Your method is correct, so if your arithmetic is right then you have the correct answer. Note, however, that when setting it out, its clearer to write what each line corresponds to! (see the red amendments)

Q:is the normal force the same as mgcos34?

Yes.

 

[Ja4Coltrane]

Now I think that the most fundamental thing that you need to have learned from this problem is how to calculate frictional force. Remember always that kinetic frictional force (friction when an object is sliding) is always equal to the coefficient of kinetic friction times the force that a surface pushes against the object (aka normal force). If you sit on a chair, the chair must push up on you with your own weight to keep you from moving. It is very important also that you understand why In this case you used cosine. The surface only pushes the object perpendicularly to the surface itself. This means that normal force is only equal to the weight of the object in the direction perpendicular to the surface.

 

[Cole07]

My answer still seems to be wrong and I've check my math a few times could someone possibly tell me what i have done wrong please?

 

[cristo]

Seems like I did read the question right the first time! You should note that this is only the work done to overcome friction. You will need to add the work done to overcome the weight of the object (i.e. the answer to part 1)

 

[Cole07]

so it would instead be
909.516396+0.30=909.816396
and then multiply this by 2.6 which would be 2365.52263J ?

 

[cristo]

No. I don't know what you've done there!

There are two ways to do this. You could use work=(frictional force+component of weight along the incline)*distance. However, since you have already calculated the work to move the box (i.e. the work to overcome the weight of the box acting down the incline) you can simple add together the work calculated in post #16 and the work calculated in question 1.

 

[Ja4Coltrane]

that should make sense if you understand the meaning of work--a change in energy. You normaly need enough work to change the potential energy from 0 to mgh, but because of friction you have to change energy from 0 to mgh to mgh+ some other amount and then frictions work gets rid of that other amount.