- #1

AdkinsJr

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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

[tex] W_{ext}=\Delta E_{mech}+f_k \Delta x [/tex]

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.

[tex]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[/tex]

Plugging in numbers and solving for the final velocity they obtain:

[tex]v_f \approx 5.6 m/s [/tex]

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:

[tex]a_x = gsin(\theta) -g \mu_k cos(\theta) [/tex]

Plugigng in numbers real fast gives:

[tex]a_x \approx 1.93 m/s^2[/tex]

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

[tex]x=\frac{1}{2}a_x t^2 [/tex]

[tex]v_f=a_x t [/tex]

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

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.