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http://xae.xanga.com/b29c717771c32184590309/b141495470.jpg 答案 c) 240 N to the left; d)240 N to the right 點解ge?? 仲有e果題 http://x90.xanga.com/7cbc5a6574430184590341/b141495480.jpg le個又點計ga....? 更新: 唔明第32題d... 點解唔係同c)一樣咁計:: F – R – f’ = ma 420 – R – (40)(2.0) = (40)(4) R = 180 N to the right

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32.a. Total frictional force, f = (30 40)(2.0) = 140.0 N By Newton’s 2nd law of motion F – f = (M m)a 420 – 140 = (30 40)a Acceleration, a = 4 ms-2 b. By Newton’s 2nd law of motion, F’ – f’ = Ma F’ = (40)(4) Net force on the 40 kg crate, F’ = 160 N c. Let R be the force exerted by the 40 kg crate on the 30 kg crate. By Newton’s 2nd law of motion, F – R – f’ = ma 420 – R – (30)(2.0) = (30)(4) R = 240 N to the left (oppose the acceleration) d. Force exerted by the30 kg crate on the 40 kg crate and that of in part c are action and reaction pairs. They are equal in magnitude but opposite in direction. They act on two and only two bodies. So, the required force = 240 N to the right e. By Newton’s 2nd law of motion, Force exerted by the 40kg crate on the 30 kg crate = (30)(4) (30)(2) = 180 N to the right So, force exerted by the 30 kg crate on the 40 kg crate, R = 180 N to the left By Newton’s 2nd law, F – R – f = Ma F – 180 – (40)(2.0) =40(4) Force, F = 420 N The worker will not find it easier to push the two crates under same acceleration. Actually, the required force is the same. 33. The horizontal component of the tension provides the force for acceleration. The vertical component balances the weight of the yoyo Tcos5? = mg T = 10m / cos5? Now, Tsin5? = ma (10m / cos5?)sin5? =ma Acceleration, a =10tan5? = 0.875 ms-2

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