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(1) 考慮 object weight 在 inclined plane 的兩個方向的 component: Parallel to the plane (平行) Perpendicular to the plane (垂直) 其中, parallel to the plane 的 component = mg sin 30° Perpendicular to the plane 的 component = mg cos 30° 所以, limiting friction 為 μsmg cos 30° 相比之下, mg sin 30° 和 μsmg cos 30°: mg sin 30°/(μsmg cos 30°) = tan 30°/μs = 1/(0.35 × √3) = 1.65 > 1 所以 mg sin 30° > μsmg cos 30°, 即個 object will fall down. (2) Since the dynamic friction coefficient is not given, we still take the static one as for calculation. Then net force (淨力) on the object = mg sin 30° - μsmg cos 30° = mg(1 - 0.35 × √3)/2 So acceleration is g(1 - 0.35 × √3)/2 = 1.97 m/s2 Finally, applying v2 - u2 = 2as with s = 2 and u = 0, we have: v2 = 2(1.97)(2) v = 2.81 m/s 這是 final velocity, 即是到 ground 時的.
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物理問題(摩擦)(an object is placed on an inclinfd plate and no extra of fix it on this place as the diagram if the (角度) lenght of this inclined plate are 30度and the 2m long repectively.coefficeient of friction (us(靜態摩擦系數))is 0.35(1)(verify wherther the object will fall from the top to... 顯示更多 物理問題(摩擦) (an object is placed on an inclinfd plate and no extra of fix it on this place as the diagram if the (角度) lenght of this inclined plate are 30度and the 2m long repectively.coefficeient of friction (us(靜態摩擦系數))is 0.35 (1)(verify wherther the object will fall from the top to groound.) 雖然知會滑落但係如何得知要用計數去求証嫁? (2)(find the velocity of the object.) 搵返個速度喎!但係我英文唔係好好嫁喳!(要列埋過程算識)最佳解答:
(1) 考慮 object weight 在 inclined plane 的兩個方向的 component: Parallel to the plane (平行) Perpendicular to the plane (垂直) 其中, parallel to the plane 的 component = mg sin 30° Perpendicular to the plane 的 component = mg cos 30° 所以, limiting friction 為 μsmg cos 30° 相比之下, mg sin 30° 和 μsmg cos 30°: mg sin 30°/(μsmg cos 30°) = tan 30°/μs = 1/(0.35 × √3) = 1.65 > 1 所以 mg sin 30° > μsmg cos 30°, 即個 object will fall down. (2) Since the dynamic friction coefficient is not given, we still take the static one as for calculation. Then net force (淨力) on the object = mg sin 30° - μsmg cos 30° = mg(1 - 0.35 × √3)/2 So acceleration is g(1 - 0.35 × √3)/2 = 1.97 m/s2 Finally, applying v2 - u2 = 2as with s = 2 and u = 0, we have: v2 = 2(1.97)(2) v = 2.81 m/s 這是 final velocity, 即是到 ground 時的.
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