How Far Will a Bundle Slide Up an Incline with Friction?

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A 4.0 kg bundle with 128 J of kinetic energy starts sliding up a 30° incline with a coefficient of friction of 0.30. Initial calculations for the normal force and friction were incorrect, leading to an estimated distance of 6.2 m. The correct normal force should be calculated as N=mgcos(30°), which adjusts the frictional force. After correcting the calculations, the distance the bundle slides up the incline is determined to be 4.3 m. Understanding the relationship between forces and drawing a free body diagram is essential for solving such problems.
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A 4.0 kg bundle starts up a 30° incline with 128 J of kinetic energy. How far will it slide up the plane if the coefficient of friction is 0.30?So, I began finding the Normal
N=W(y)=4cos30=3.46

Then found friction

f=uN
f=.3*3.46=1.04

So
E=E
K=Ug + Q
K=mgh + fd
K=mg(dsin30)+fd
128=4(9.8)(.5)d+(1.04)d
128=20.64d
d=6.2m

But the answer says 4.3m. What's wrong?

Thanks!

EDIT:
So I figured it out... N=4*9.8cos30. Duh.
 
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h7u9i said:
So I figured it out...
Glad things worked out. By the way, welcome to Physics Forums!
 
I had a similar problem (just the numbers changed). I get the whole h = dsin(30) thing, but why does Fn = mgcos(30)?
 
Ghostscythe said:
I had a similar problem (just the numbers changed). I get the whole h = dsin(30) thing, but why does Fn = mgcos(30)?
Draw a free body diagram. Take the direction of the normal force as the y-direction, apply Fnet=ma to the y-direction, and solve for Fn.
 
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