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151_Physics ProblemsTechnical Physics

151_Physics ProblemsTechnical Physics - Chapter 5*P5.70 153...

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Chapter 5 153 *P5.70 Throughout its up and down motion after release the block has F ma n mg n mg y y = + = = : cos cos . θ θ 0 Let R i j = + R R x y ± ± represent the force of table on incline. We have F ma R n R mg F ma Mg n R R Mg mg x x x x y y y y = + = = = + = = + : sin cos sin : cos cos . θ θ θ θ θ 0 0 2 R = + + mg M m g cos sin cos θ θ θ to the right upward 2 e j FIG. P5.70 *P5.71 Take + x in the direction of motion of the tablecloth. For the mug: F ma a a x x x x = = = 0 1 0 2 0 5 . . . . N kg m s 2 Relative to the tablecloth, the acceleration of the mug is 0 5 3 2 5 . . m s m s m s 2 2 2
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