Incline problem with friction; alternative solution

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SUMMARY

The discussion revolves around solving an incline-mass problem involving friction, specifically a 40 kg mass on an 8 m slide at a 30-degree angle with a coefficient of kinetic friction of 0.35. The correct final speed at the bottom of the slide, calculated using work-energy principles, is approximately 5.6 m/s. The user attempted to solve the problem using Newton's second law and kinematic equations but mistakenly used an incorrect slide length of 80 m, leading to an erroneous final speed of 17.6 m/s. The error was identified as a simple oversight rather than a misunderstanding of the underlying physics concepts.

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Homework Statement


There is an example problem in a textbook I'm looking at where they solve a simple incline-mass problem (friction included) using work-energy.

We are given the mass 40kg, the length of the slide is 8 m, the incline is 30 degrees, the coefficient of kinetic friction is .35. We are looking for the child's speed at the bottom of the slide.

Homework Equations


W_{ext}=\Delta E_{mech}+f_k \Delta x

Newton's Seconds Law, the basic kinematic equations:

The Attempt at a Solution



The solution in the example is straight forward. The child-slide-earth system has no exterior forces acting so the equation I give above is set to zero.

0=mg\Delta h +\frac{1}{2}mv_f^2+f_k \Delta x=-mg\Delta x sin(\theta)+\frac{1}{2}mv_f^2 +mg\mu_k cos(\theta)\Delta x

Plugging in numbers and solving for the final velocity they obtain:
v_f \approx 5.6 m/s

I am attempting the same problem usuing Newton's second law and kinematic equations, but I am not obtaining a similar answer and I can't discern why not:

In the x-direction, using the usual coordinate system (x - axis parallels to slope of the incline, +x is down the incline), the acceleration should be:

a_x = gsin(\theta) -g \mu_k cos(\theta)

Plugigng in numbers real fast gives:
a_x \approx 1.93 m/s^2

Now I attempt to use kinematics to find the speed at the bottom of the slide:

x=\frac{1}{2}a_x t^2

v_f=a_x t

80m = \frac{1}{2}(1.93 m/s^2)t^2\Rightarrow t\approx 9.11s

Using this in the velocity equation along with the acceleration of 1.93 m/s^2 gives about 17.6 m/s for the speed. This is far different than the 5.6 m/s obtained using the energy equations.The numbers I obtained don't pass sanity check either.
 
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In your last equation you've used 80 m as the length of the slide. Wasn't the value only 8 m in the problem statement?
 
gneill said:
In your last equation you've used 80 m as the length of the slide. Wasn't the value only 8 m in the problem statement?

You got to be kidding me. I can't believe I didn't notice that and wasted this much time. I guess I'm kind of relieved that it was my vigilance and not my understanding of physics that failed though. I thought I was going crazy for moment.
 
Last edited:

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