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

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SUMMARY

A 4.0 kg bundle with an initial kinetic energy of 128 J slides up a 30° incline with a coefficient of friction of 0.30. The normal force was initially miscalculated, leading to an incorrect distance of 6.2 m instead of the correct 4.3 m. The correct normal force is calculated using N = mgcos(30°), which is essential for determining the frictional force and the distance traveled up the incline. A free body diagram is recommended for visualizing forces acting on the bundle.

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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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