Space Time & Motion A.2 Forces & Momentum (id: 7ca26aafb)

本题目来源于试卷: Space Time & Motion A.2 Forces & Momentum，类别为 IB物理学

[填空题]
1.Define Newton's Thipo9 lyu ufs,z;u5 0l:rrd Lyro+8l gl8st h0y s70baw.
2.A block of mass 2.0 kg is placed on another block of mass 5.0 kg. The blocks sit on a frictionless surface, but the coefficient of static friction between the blocks is 0.50. The block at the top is pulled with a varying horizontal force F


2.1Calculate the magnitude of the maximum force $F_{max}$ which results in both of the blocks moving together without slipping.  N
2.2On the graph, sketch how the force of static friction $F_f$ acting on the bottom block varies with the force F
(no need to add values on the axis)



$F_{max}$ is your answer to (b)(i) and draw your graph up to $F_{max}$
2.3According to Newton's Third Law, there must be another force paired with the weight of the top block. Describe this force.
3.On another occasion, the blocks are moved with the help of a force of 12N acting on the block at the top. The drag force D on the blocks is given by D=2.5*$10^{-2}v^2$where v is the speed of the blocks. Calculate the top speed of the blocks for this force.  $ms^{-2}$

参考答案:
空格1: 14  空格2: 22 


本题详细解析:

1. when two bodies A and B interact, the $\ll$action$\gg$ $\underset{‾}{\text{force}}$ that A exerts on B is equal and $\mathrm{✓}$

   opposite in direction to the $\ll$reaction$\gg$ $\underset{‾}{\text{force}}$ that B exerts on A $\mathrm{✓}$

2. 1. The magnitude of maximum force of friction = ${\mu }_{s}R=9.8N\mathrm{✓}$

      The acceleration of both blocks = $\ll \frac{F_{fmax}}{5}=\gg 2.0m{s}^{-2}\mathrm{✓}$

      $F=ma=\ll (5+2)(5)=\gg$

      $14N$
      $\mathrm{✓}$

      $\underset{‾}{\text{Further Explanation}}$: 
      To have the blocks moving together, the acceleration of each block must be the same. The vertical forces on each block cancel each other. Therefore, only horizontal forces create acceleration. The only horizontal force acting on the block with a mass of $5.0kg$ is the force due to static friction between the blocks. The acceleration of $5.0kg$ mass can be found by Newton's second law ($F_{net}=ma$);

      

      The acceleration of $5.0kg$ mass can be found by Newton's second law ($F_{net}=ma$);

      $F_f=5a$ (first expression)

      Two horizontal forces are acting on the 2.0 $kg$ mass as $F$ and the force of static friction.

      

      The acceleration of $2.0kg$ mass can be found by

      $F-F_f=2.0a$ (second expression)

      

      
      

      

      

      
      

      The acceleration of the blocks must be zero when they are moving at top speed. The resultant force should be zero to have no acceleration. Since two forces acting on the blocks are drag force and 12 N, they must be balanced by each other;

      
      

      

      
      

      
      

      
      

      
      

      
      

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