# How do I find the work done by the kinetic friction force when the surface is a curve

#### Chemist116

The problem is as follows:

The diagram from below shows a boy of 30 kg of mass slides down a slide from a height of $5\,m$ starting from rest in point $A$. He reaches to point $B$ with a speed of $4\,\frac{m}{s}$. Find the work done by the frictional force in Joules. You may use $g=9.81\,\frac{m}{s}$.

$\begin{array}{ll} 1.&-981.5\,J\\ 2.&-1231.5\,J\\ 3.&-1421.5\,J\\ 4.&-1551.5\,J\\ 5.&-1980.5\,J\\ \end{array}$

I'm really totally lost on this one. What I attempted to do was to use the conservation of mechanical energy. But I don't see how it could be applied here.

I assumed that the Kinetic energy is what he achieves in the bottom as:

$E_{k}=\frac{1}{2}mv^2$

and the potential energy he had in the top is:

$E_{u}=mgh$

Therefore since there is a conservation both expressions are the same:

$\frac{1}{2}mv^2=mgh$

But here's where I' lost as I'm given the mass, the height and the speed. I don't seem to find the earlier equation of any use. What would be right method to use to solve this problem?. Can somebody help me here? and more importantly why the work is negative here?. :help:

#### romsek

Math Team
At the top the kid has no kinetic energy and $mgh_i$ in potential energy

At the bottom it has $\dfrac{mv^2}{2}$ in kinetic energy and $mgh_f$ in potential energy

All of these can be calculated with the data given.

The work done by the frictional force is the difference between the total energy of the end final and initial states, i.e.

$(KE_f+PE_f) - (KE_i + PE_i) = \\ \left(\dfrac{m v^2}{2}+ mgh_f\right) - \left(0+ mgh_i \right) = \\ \dfrac{m v^2}{2} + mg(h_f - h_i)$