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Grace asked in 科學其他:科學 · 1 decade ago

物理題目~英文唷~幫我解題

第一題

A 3.0-kg block is on a horizontal surface. The block is at rest when, at t = 0, a force (magnitude P = 12 N) acting parallel to the surface is applied to the block causing it to accelerate. The coefficient of kinetic friction between the block and the surface is 0.20. At what rate is the force P doing work on the block at t = 2.0 s?

A. 54 W B. 49 W C. 44 W D. 59 W E. 24 W

第二題

A 20-kg block on a horizontal surface is attached to a light spring (force constant = 8.0 kN/m). The block is pulled 10 cm to the right from its equilibrium position and released from rest. When the block has moved 2.0 cm toward its equilibrium position, its kinetic energy is 13 J. How much work is done by the frictional force on the block as it moves the 2.0 cm?

A. –2.5 J B. –1.4 J C. –3.0 J D. –1.9 J E. –14 J

第三題

A 0.80-kg object tied to the end of a 2.0-m string swings as a pendulum. At the lowest point of its swing, the object has a kinetic energy of 10 J. Determine the speed of the object at the instant when the string makes an angle of 50 with the vertical.

A. 5.6 m/s B. 4.4 m/s C. 3.3 m/s D. 5.0 m/s E. 6.1 m/s

需要計算過程

Update:

可以用中文

嘛我看不懂

3 Answers

Rating
  • 1 decade ago
    Favorite Answer

    順便翻譯題目^^

    第一題

    水平表面上有一3kg的block, 有一平行表面定力作用在block上

    且大小是12N,平面的動摩擦係數(coefficient of kinetic friction)

    是0.2,在 t = 2 sec 時此定力作用於block的順時功率是多少?

    A. 54 W B. 49 W C. 44 W D. 59 W E. 24 W

    解:

    瞬時功率 = F * v 所以得先算出 t=2 時的速度v

    由 F = ma 合力 = 12- 3*9.8*0.2 = 6.12 = ma = 3 * a

    所以 a = 2.04

    t = 2 時,順時速度 v = at = 2.04*2 = 4.08

    最後由瞬時功率 = F * v 功率 = 12 * 4.08 = 48.96

    近似值 = 49 W

    答案B

    打一題花不少時間^^",二三題我明天晚上在打,雖然我

    只是初學者五級,但你可以放心^^,我是物理系畢業的

    ,先跟你講第二第三題答案是B和C^^

    2007-11-17 17:50:44 補充:

    第二題

    水平面上有一block質量20kg,綁在一彈力常數是8k,也就是8000的

    輕彈簧上,這一題是在問能量守恆,原本的力學能 = 後力學能+熱能

    所以算式是 原本的彈力位能 = 最後的彈力位能 + 動能 + 摩擦力作功

    注意彈簧伸長量要換算成公尺 10cm=0.1m ,8cm=0.08m

    即 0.5 * 8000 * (0.1的平方) = 0.5 * 8000 * (0.08的平方) +13 + 熱能

    摩擦力造成的熱能 = 40 - 25.6 -13 =1.4

    題中答案帶負號是因為 摩擦力和位移方向相反,摩擦力做負功

    所以答案是 -1.4J ...........(B)

    2007-11-17 17:50:56 補充:

    第三題

    有一個單擺,擺球質量0.8kg,擺長2.0m,在球在最低點時動能是10

    J,問單擺跟鉛直面夾50度角時,球的速度是多少

    力學能守恆的題目 假設最低點為位能是0的地方的話

    初動能 + 位能 = 末動能 +位能

    10 + 0 = 末動能 + 0.8 * 9.8 * (2 - 2cos50度)

    末動能 = 10 - 5.60 = 4.40

    4.40 = 0.5 * m * v平方 = 0.5 * 0.8 * v平方

    v = 3.3 (m/s)

    這題的計算得按工程計算機,cos50度數字很醜

    2007-11-17 17:51:22 補充:

    補:第一題到底要求瞬時功率還是平均功率,我的判斷標準是看

      英文,題目寫at ,所以我覺得是在 t=2 時的功率,也就是瞬

      時功率,如果要算平均功率的話,bdref43大大也給答案了,

      如果看不懂的話再說吧

      bdref43 大大好銷魂的學歷,難怪解題用全英文,不過原po應

    該是小大一,在俢普通物理學吧,看原文會抖的,我大一也

      是一樣,題目改成中文,你大概就會算了,抱歉一直補充內

      容喔^^"

    Source(s): 自己, 自己
  • 1 decade ago

    高手!!

  • ?
    Lv 7
    1 decade ago

    1. Assuming g=10 m/s2, the friction force=mgu=3.0 kg*10 m/s2*0.20=6.0 N. The net force applied to the block is 12-6.0=6.0 N, producing acceleration at 6.0/3=2.0 m/s2. At t=2 the block has traveled at2/2 or 4.0 m. The rate of the work done is 12*4.0/2=24 W. Ans. (E)

    2. The potential energy stored in the string when it is pulled is kX2/2 or 8000*0.12/2=40 J. When it moved 2.0 cm, the potential energy changes to 8000*0.082/2=25.6 J. 40-25.6=14.4 J was converted but the kinetic energy was only 13 J. Therefore, 13-14.4=-1.4 J was due to the work done by the friction force. Ans. (B)

    3. Let the spring constant be K and the elongation at the lowest point be S. Let g be 10 m/s2 for convenience. So, at the lowest point, mg=kS, i.e. 0.8*10=kS=8, S=8/k. The potential stored is kS2/2=32/k. When the string makes 50 degree angle, the balance of the force becomes mgsin(50)=kS'=0.8*10*0.766=6.1, i.e. the enlongation becomes 6.1/k. The difference in height of the block is dh=(2+8/k)-(2+6.1/k)*cos(50)=0.72+3.9/k. The change in potential energy is 8*dh=5.8+31.2/k J. From energy conservation, 10+32/k=5.8+31.2/k+E where E is the kinetic energy at the 50 degree angle. Ignoring 0.8/k, E=4.2=mV2 0.8*V2/2, V=3.3 m/s

    2007-11-17 00:40:07 補充:

    3. Ans. (C)

    2007-11-17 10:28:23 補充:

    1. 題目是否問瞬時功率並不是很明顯,上面給的是2秒內的平均功率,瞬時功率為12*2*2=48 W, 取近似值B.

    2007-11-17 21:49:54 補充:

    1. 假設 g=10 m/s^2, 摩擦力=mgu=3.0 kg*10 m/s^2*0.20=6.0 N. 施於物塊的淨力為12-6.0=6.0 N, 所產生的加速度為 6.0/3=2.0 m/s^2. 在 t=2時物塊位移為 at^2/2即 4.0 m. 平均功率是12*4.0/2=24 W. 答 (E)

    2007-11-17 21:50:32 補充:

    2. 彈簧被拉後所儲存的位能為 kX^2/2,即 8000*0.12/2=40 J. 當它移動2.0 cm後, 儲存的位能變成 8000*0.082/2=25.6 J. 40-25.6=14.4 J的能量被轉換而動能只有 13 J. 因此, 13-14.4=-1.4 J被磨擦力作功消耗了. Ans. (B)

    2007-11-17 22:04:34 補充:

    3. 令彈力係數為K,彈簧在最低點的伸長量為S, g為10 m/s^2. 所以在最低點時 mg=kS,即 0.8*10=kS=8, S=8/k. 彈簧的位能為kS^2/2=32/k. 當彈簧與法線成50度時, 彈簧線上力的平衡變成 mgsin(50)=kS'=0.8*10*0.766=6.1, 即彈簧伸長量變成 6.1/k. 物體上升的高度是 dh=(2+8/k)-(2+6.1/k)*cos(50)=0.72+3.9/k. 造成位能的改變為8*dh=5.8+31.2/k J.

    2007-11-17 22:04:52 補充:

    從能量守恆得10+32/k=5.8+31.2/k+E 其中E是物體成50度角的動能. 0.8/k太小可以忽略, 所以E=4.2=mV^2/2=0.8*V^2/2, V=3.3 m/s.

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