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So I am studying for my exam on monday for analytical mechanics. I am kind of stuck on a question, so wondering if someone can give me a pointer. I am actually looking for some pointers and some techniques (hints and tips) to get me through this. the topic is just simple velocity as a function of time, velocity as a function of displacement, quadratic air resistance, oscillations.

So here is the question:

A block of wood is projected up an inclined plane with initial speed $v_0$. If the inclination of the plane is $30^\circ$ and the coefficient of sliding friction is $\mu_k = 0.1$, find the total time for the block to return to the point of projection.

Heres my work: $$F_{\text{net}} = -F_f = -\mu_k \cdot m \cdot g \cdot sin(30) = m \cdot \frac{dv_x}{dt}$$ so the m's cancel and solving the differential equation you get the following: $$ \int_{v_0}^v dv = \int_0^t -\mu_k \cdot g \cdot sin(30) dt$$ $$v-v_0 = -\mu_k \cdot sin(30) \cdot g\cdot t$$ so $$ v(t) = v_0 - 0.98\cdot sin(30) \cdot t $$

so if i set that to 0 i know the time when it reaches the top and starts to slide backwards. Now how would I know when it reaches the point of projection? Would it just be double the time?

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closed as off-topic by tpg2114, Emilio Pisanty, Qmechanic Nov 10 '13 at 2:44

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2 Answers

You've forgotten to include the force of gravity. When you do, you'll find that the time to come to a rest going up the plane is shorter than the time to go back down. That's because as the block goes up the plane, both gravity and friction are pushing it back down, but as the block comes down the plane, gravity is pulling it down while friction is trying to push it back up.

For the same reason, if you throw a ball up in the air, it takes longer to come down than to go up.

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This is an algebra physics problem, no need for calculus.

Gravity force (in the direction of sliding) is $mg\sin(30)$. Friction force is therefore $0.1mg\cos(30)$ and opposes velocity. From these you get the accelerations (different for going up and down). Now move the block up to its turning point using the going up acceleration (which is the one with the largest magnitude), and from its position, figure out the time to get back down using the coming down acceleration.

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