Incline problem with friction; alternative solution

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The discussion revolves around solving a physics problem involving a child sliding down an incline with friction. The original solution using work-energy principles yields a final speed of approximately 5.6 m/s. An attempt to solve the problem using Newton's second law and kinematic equations resulted in an incorrect speed of about 17.6 m/s due to a miscalculation involving the length of the slide, mistakenly using 80 m instead of the correct 8 m. The participant expresses relief upon realizing the error was a simple oversight rather than a misunderstanding of the physics concepts. This highlights the importance of careful attention to detail in problem-solving.
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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:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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