Frictional force and work problem

AI Thread Summary
The problem involves calculating the magnitude of the frictional force acting on a woman sliding down a 35-meter high slope inclined at 45 degrees, with a given work done by friction of -8.58 x 1000 J. The correct approach requires using the hypotenuse of the slope for distance rather than the height, which is calculated as 35/sin(45). The frictional force can be determined using the formula F = Work done / distance, leading to a calculated force of approximately 173 N. The initial velocity of the woman at the bottom does not impact the calculation of frictional force in this context. Understanding the direction of the frictional force is crucial for accurate calculations.
billu77
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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
 
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billu77 said:

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
 
Hi billu77! :wink:
billu77 said:
-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 … ? :smile:
 
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.
 
caitlincui said:
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
 
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