Calculate the work done by the kinetic frictional force

[Student3.41]

Homework Statement

I have a few questions I need help with on my assignment. I cannot wrap my head around this work and energy chapter.

Anyway here it is;

A 62.7-kg skier coasts up a snow-covered hill that makes an angle of 25.2° with the horizontal. The initial speed of the skier is 8.22 m/s. After coasting 2.09 m up the slope, the skier has a speed of 3.20 m/s. Calculate the work done by the kinetic frictional force that acts on the skis

Homework Equations

W=Fd
Wtotal=1/2mvf^2-1/2mvi^2

The Attempt at a Solution

FBD. Fn in the positive y-direction, mgcos(theta) in the negative y-direction, Fk and mgsin(theta) in the negative y-diretion.

W_FN=0 ----> makes a 90deg angle with the direction.
W_mgcos(theta)=0

W_mgsin(theta)=Fd
=(mgsin25.2)(2.09m)
=(62.7kg)(9.8m/s^2)(0.426)(2.09m)
=548J
W_Fk= ?

W_TOT= 0+0+548+W_Fk = 1/2mVf^2-1/2mVi^2

=> 548J+W_Fk=321J-2120J
=> W_Fk=-1800J-548J=-2350J

I am using three sig digits for my calculations. The answer is wrong.

 

[tiny-tim]

Hi Student3.41!

(your equations would be easier to read ifyouleftsomespaces)

W_TOT= 0+0+548+W_Fk = 1/2mVf^2-1/2mVi^2

no, your LHS is positive, so your RHS should be positive also …

interchange V_f and V_i

 

[Student3.41]

Hi Student3.41!

(your equations would be easier to read ifyouleftsomespaces)


no, your LHS is positive, so your RHS should be positive also …

interchange V_f and V_i

OK, so interchange the values so;

545J+W_K= 1/2mV_i^2-1/2mV_f^2

= 2120J-322J
=1780J

W_K= 1780J-545J = 1240J

But this is wrong. Unless I am not understanding what you mean. Only have 3 more trys :O

 

[tiny-tim]

W_K= 1780J-545J = 1240J

But this is wrong. Unless I am not understanding what you mean. Only have 3 more trys :O

isn't it 548, not 545?

 

[Student3.41]

isn't it 548, not 545?

I re calculated my answer so I didn't take everything by 3 significant digits. Either way the answer comes out to be wrong.

 

[tiny-tim]

I re calculated my answer so I didn't take everything by 3 significant digits. Either way the answer comes out to be wrong.

Sorry, then I can't see what's wrong.

(btw, you should keep everything to at least 4 sig figs until the very end, and only then round off to 3 sig figs; and I made it 547.4)

 

[Student3.41]

Sorry, then I can't see what's wrong.

(btw, you should keep everything to at least 4 sig figs until the very end, and only then round off to 3 sig figs; and I made it 547.4)

I am not sure if that will affect the answer, my prof usually gives us some room for error.

I am not sure either, maybe an error in the answer.

Isn't the Work of friction always -fk because W_fk = fkcos180d = fk(-1)d = -f_kd?

Also, is there a need to move the 547.7 to the RHS?

I am also posting another problem if you can help with that! I have done all the calculations and its been driving me crazy.. Thanks for the help :)

 

[tiny-tim]

Isn't the Work of friction always -fk because W_fk = fkcos180d = fk(-1)d = -f_kd?

Yes, the work done by both friction and (in this case) gravity will be negative because those forces are in the opposite direction to the https://www.physicsforums.com/library.php?do=view_item&itemid=378"

 

[Student3.41]

Yes, the work done by both friction and (in this case) gravity will be negative because those forces are in the opposite direction to the https://www.physicsforums.com/library.php?do=view_item&itemid=378"

I got the right answer. All the workings were correct. I was not putting in the (-) sign before my answer. Came out to be -1250J