紅茶 asked in 社會與文化語言 · 1 decade ago

非常緊急 徵高中物理小老師

<1>A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6.43). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s. (a) What is the tension in the cord in the original situation when the block has speed v = 0.70 m/s? (b) What is the tension in the cord in the final situation when the block has speed v = 2.80 m/s? (c) How much work was done by the person who pulled on the cord?

2.Crash Barrier. A student proposes a design for an automobile crash barrier in which a 1700-kg sport utility vehicle moving at 20.0 m/s crashes into a spring of negligible mass that slows it to a stop. So that the passengers are not injured, the acceleration of the vehicle as it slows can be no greater than 5.00 g. (a) Find the required spring constant k, and find the distance the spring will compress in slowing the vehicle to a stop. In your calculation, disregard any deformation or crumpling of the vehicle and the friction between the vehicle and the ground. (b) What disadvantages are there to this design?

2 Answers

Rating
  • ?
    Lv 7
    1 decade ago
    Favorite Answer

    (1) 一個0.120 kg的小塊由一條通過一個無磨擦力平面上的一個小洞的繩子綁著在平面作圓周運動.起先小塊離小洞0.40m時速率為0.70m/s,現將繩子往下拉使圓周半徑縮小成0.10m,此時小塊速率為2.80m/s.問(a)繩子最初的張力, (b)最後的張力, (c)繩子拉下時作功多少?

    (a) 圓周半徑為0.40m, V=0.70=rw, 角速度 w=0.70/0.40=1.75, 離心加速度 a=rw2=0.40*1.752=1.225, 繩子張力=離心力, T=F=ma=0.12*1.225=0.147 N

    (b) 圓周半徑為0.10m, V=2.80=rw, 角速度 w=2.80/0.10=28.0, 離心加速度 a=rw2=0.10*28.02=78.4, 繩子張力=離心力, T=F=ma=0.12*78.4=737.6 N

    (c) 作功=小塊的動能變化=0.120*(2.802-0.702)/2=0.44 J.

    (2) 一個學生提出一個攔阻牆的設計,可以使一部1700 kg的車子以20.0 m/s的速度撞上時,牆裡的一個彈簧(不計質量)能使車子停止而不傷及人員.但減速度不能超過5g. (a)不計彈簧壓縮時變形以及車子變形和與地面的摩擦力,求彈簧的彈力係數, (b) 這個設計的缺點何在?

    (a) 由能量不滅知車子動能=彈力位能,要以5g的減速度停止(g=9.8 m/s2), 彈簧必須被壓縮的距離為 X=(0-20.02)/(2*5g)=-4.08 m, mV2/2=kX2/2, k=m(V/X)2=1700*(20.0/(-4.08))2=40800 N/m

    (b) 要做出有這麼大的彈力係數的彈簧非常不容易.

  • Anonymous
    6 years ago

    到下面的網址看看吧

    ▶▶http://qaz331.pixnet.net/blog

Still have questions? Get your answers by asking now.