Solving Work Needed to Push 1147kg Car Up 15.1o Incline

[Jayhawk1]

I am having a lot of trouble with this problem:

What is the minimum work needed to push a 1147 kg car 376 m up a 15.1o incline if the effective coefficient of friction is 0.23?

I thought that it was (m)(.23)(sin 15.1)(376) but that isn't right. Any help would be appreciated.

 

[dextercioby]

Nope,my guess is that it should encompass a "cosine".So what is the force up against whom u have to do work...?

Daniel.

 

[Jayhawk1]

So it should be (m)(.23)(cos15.1)(376)? ...for some reason I though the coeff. of friction incorporated the angle... am I wrong in assuming this?

 

[Nenad]

Just use the work energy theorem.
\Delta W = \Delta E_g + \Delta E_f + \Delta E_s

Draw a free body diagram first, it will help you.

Regards,

Nenad

 

[dextercioby]

There are 2 forces agains which work needs to be performed:gravity & friction force...
You have found the one for gravity (it should be with a minus) and u need to add to it the work performed by friction force.

Daniel.

 

[marlon]

Suppose the x-axis is along the incline and y-axis is perpendicular to this surface...


We have 3 fources on the object : gravity, friction{\mu}{\vec N}
, normal force N

So along the x-axis we have :

F_x = -mgsin(\theta) - {\mu}N

Along the y-axis we have :


F_y = 0 = -mgcos(\theta) + N



Now, you can calculate N from the second equation and then plug it into the first one. Now you know everything along the incline (x-axis). So just fill in the numbers and multiply this by the distance that is travelled.

regards
marlon

 

[Jayhawk1]

So for the component of force due to friction it is (coeff. friction)(mass)(376)?

 

[marlon]

There are 2 forces agains which work needs to be performed:gravity & friction force...
You have found the one for gravity (it should be with a minus) and u need to add to it the work performed by friction force.

Daniel.

Wrong, besides you confused the OP, when you started about "encompassing" a cosine...his formula was initially correct for the gravity-part.

marlon

 

[marlon]

So for the component of force due to friction it is (coeff. friction)(mass)(376)?

Nope, read my post...

it is (coeff friction)(NORMAL FORCE) for the formula of the friction force along the incline. Beware of the minus-sign. Then multiply this by 376 to get the work done...

I showed you how to calculate the normal force N

marlon

 

[dextercioby]

What's wrong in my statement that u quoted...? :

Daniel.

 

[marlon]

What's wrong in my statement that u quoted...? :

Daniel.

Look at post number 3 !

marlon

 

[dextercioby]

You said "wrong" when u quoted something right,isn't that so...?

Marlon,what can i say more...?

Daniel.

 

[marlon]

You said "wrong" when u quoted something right,isn't that so...?

Marlon,what can i say more...?

Daniel.

Please, stop being such a big baby about this. You know very well you are wrong. Now, you don't have to admit it to me, but just stop posting useless answers just to have the last word.

marlon

Stick to the facts

 

[marlon]

F_x = -mgsin(\theta) - {\mu}mgcos(\theta)

I just replaced N with the second equation. Now for the work, you can omit the minus-signs

Fill in the numbers and multiply by the traveled distance and you are done. ps disregard dexter's posts because they are erronuous...

marlon