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

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Homework Help Overview

The discussion revolves around a physics problem involving a bundle sliding up an incline with friction. The scenario includes a 4.0 kg bundle on a 30° incline, starting with 128 J of kinetic energy, and a coefficient of friction of 0.30. Participants are exploring the calculations related to the distance the bundle will slide up the incline before coming to a stop.

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Approaches and Questions Raised

  • One participant attempts to calculate the distance using energy conservation principles, but questions arise regarding the normal force and friction calculations. Another participant seeks clarification on the relationship between the normal force and gravitational components, specifically why Fn = mgcos(30).

Discussion Status

The discussion includes attempts to clarify calculations and concepts related to forces on an incline. Some participants have provided guidance on drawing free body diagrams and applying net force equations, while others are exploring similar problems and seeking deeper understanding of the underlying principles.

Contextual Notes

There is a mention of a discrepancy in the calculated distance, with one participant noting a difference between their result and the expected answer. The discussion also reflects on the importance of correctly determining the normal force in the context of the problem.

h7u9i
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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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