Frictional force and work problem

[billu77]

Homework Statement

A slide similar to der stuka is 35.0 meters high but is a straight slope,inclined at 45.0 degrees with respect to the horizontal. If a 60kg woman has the speed of 20.0m/s at the bottom and the work done by frictional force on the woman is -8.58 x 1000J, find the magnitude of the force of friction.

Homework Equations

Work done by frictional force = Frictional force cos 45 x horizontal distance

The Attempt at a Solution

using the equation:
-8580 = F cos 45 x 35

i am getting double the correct answer...can u tell me what am i doing wrong

 

[Andrew Mason]

Homework Statement

A slide similar to der stuka is 35.0 meters high but is a straight slope,inclined at 45.0 degrees with respect to the horizontal. If a 60kg woman has the speed of 20.0m/s at the bottom and the work done by frictional force on the woman is -8.58 x 1000J, find the magnitude of the force of friction.

Homework Equations

Work done by frictional force = Frictional force cos 45 x horizontal distance

The Attempt at a Solution

using the equation:
-8580 = F cos 45 x 35

i am getting double the correct answer...can u tell me what am i doing wrong

You have to divide the height by cos 45

AM

 

[tiny-tim]

Hi billu77!

-8580 = F cos 45 x 35

i am getting double the correct answer...can u tell me what am i doing wrong

yes … work done = force times distance, the height is 35, so the distance along the slide is … ?

 

[caitlincui]

Your trig function is wrong. W=Fdcos(0), since the work the friction done is 8580 J, then W=8580. The distance the friction gone through is Hypotenuse, which is 35/sin(45) rather than 35*cos(45). F=8580/(35/sin(45))=8580*sin(45)/35=173. I am not sure what the initial velocity is doing though.

 

[billu77]

Your trig function is wrong. W=Fdcos(0), since the work the friction done is 8580 J, then W=8580. The distance the friction gone through is Hypotenuse, which is 35/sin(45) rather than 35*cos(45). F=8580/(35/sin(45))=8580*sin(45)/35=173. I am not sure what the initial velocity is doing though.

thanks...i did not pay attention to the fact that frictional force is just acting along the slide and not at an angle...appreciate ur help