capa7 - 1 This problem is related to last-weeks capa...

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1) This problem is related to last-week’s capa problem 8: Sliding down at constant speed means that the net-force is zero. Or in other words: N + m g + F friction = 0 or with the magnitudes: F friction - mg·sin θ =0 (the second term is the component of m g + N downward along the slope). mg N v F If the object is sliding upwards, mg·sin θ and F both point downwards along the slope. The net-force in this case is no longer zero but rather F friction + mg·sin θ = 2·mg·sin θ F net = ma b a = F net /m =2mg· sin θ / m = 2g·sin θ Since the acceleration is constant we can use the eqn for constant acceleration: v = v 0 – a t (‘ – ,’ because the acceleration points in the opposite direction of the direction of motion.) ‘… the object comes to rest ‘ means that v = 0 b time ‘t’ until the object comes to rest: t = v 0 / a = v 0 / (2g·sin θ ) (here we used the expression for ‘a’ from above.) Under constant deceleration the change in position is given by: x = x 0 + v 0 t – ½ a t 2 . As often done before, we chose the x-axes in a way that x 0 = 0. Using the expressions for ‘t’ and ‘a’ from above we find: x = v 0 2 /a – ½ a v 0 2 /a 2 = ½ v 0 2 /a = v 0 2 / (4g sin θ ) 2) The work ‘W’ done by the force F along the path Δ r is defined as: W = F · Δ r = | F |·| Δ r |·cos θ , where θ is the angle between F and Δ r . In the example provided here, the net force F
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This note was uploaded on 11/07/2010 for the course PHYS 1120 taught by Professor Rogers during the Spring '08 term at Colorado.

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capa7 - 1 This problem is related to last-weeks capa...

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