# Incline problem with friction; alternative solution

1. Oct 15, 2016

1. The problem statement, all variables and given/known data
There is an example problem in a text book 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.

2. Relevant equations
$$W_{ext}=\Delta E_{mech}+f_k \Delta x$$

Newton's Seconds Law, the basic kinematic equations:

3. 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.

2. Oct 15, 2016

### Staff: Mentor

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?

3. Oct 15, 2016