Calculating Frictional Work on a Child Sliding Down a Playground Slide

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To calculate the work done by friction as a child slides down a playground slide, start with the energy conservation equation: initial potential energy plus work equals final kinetic energy plus thermal energy. Given that the initial kinetic energy and final potential energy are both zero, the equation simplifies to initial potential energy plus work equals final kinetic energy plus thermal energy. The initial potential energy is 367 J, and the final kinetic energy is 187 J. By assuming that the work done by friction encompasses both the work and thermal energy, the relationship can be established to find the frictional work.
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A 40 kg child climbs a 2.5 m vertical ladder to the top of a playground slide. Starting from rest at the top of the ladder, the child slides down the incline, which makes an angle of 22 degrees with the horizontal ground. Friction is present during the descent and the child reaches the bottom of the slide with a speed of 3 m/s.

Determine the work done by friction as the child comes down the slide.

I know that the P(i) = 367 J and K(f) = 187 J. What should I do next?
 
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You should come up with a relationship between work and energy, including work lost by friction. From there it's a cinch from what you already have.
 
Well, from our energy conservation equation, P(i) + K(i) + W = P(f) + K(f) + Thermal Energy
In this case, there is no initial kinetic energy or final potential energy, so our equation would be P(i) + W = K(f) + Thermal Energy, work being the friction force. But then there are two unknowns, both the work and the Thermal energy?
 
Assume that friction work and thermal energy go hand in hand, in other words the frictional "work" term encompasses them both.
 
Oh, I get it. Thanks :)
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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