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Or using the work-energy equation...$$(F_I{_I} + F_F) d cos\theta = \frac{1}{2} m\cdot v_f^2$$Or using the principle of conservation of energy...$$ W = mv_f$$In summary, the biker will lose potential energy due to gravity and gain kinetic energy when rolling down an incline.

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It looks right, but a bit overcomplicated IMO. One good thing about a free-body diagram is that you can label the forces with an arrow showing the direction of each force. This allows you to jump in with the algebra already simplified somewhat. If ##f## is the friction force acting up the slope and ##l## is the length of the slope, then you can simply write$$F_{net} = mg \sin \theta - f$$or$$W_{net} = (mg \sin \theta - f)l$$You don't have to invoke angles of ##\phi =0## and ##\phi = \pi## etc.paulimerci said:

There's something to be said for keeping things simply initially until you develop some intuition for these problems.

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Thank you; your approach looks simple and easy to understand.PeroK said:It looks right, but a bit overcomplicated IMO. One good thing about a free-body diagram is that you can label the forces with an arrow showing the direction of each force. This allows you to jump in with the algebra already simplified somewhat. If ##f## is the friction force acting up the slope and ##l## is the length of the slope, then you can simply write$$F_{net} = mg \sin \theta - f$$or$$W_{net} = (mg \sin \theta - f)l$$You don't have to invoke angles of ##\phi =0## and ##\phi = \pi## etc.

There's something to be said for keeping things simply initially until you develop some intuition for these problems.

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