Space Time & Motion A.2 Forces & Momentum (id: f0f131c31)

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]
Two blocks of masses 3.0 kg and 4.0kg are held stationary with a thread connecting the blocks. The mass of 3.0 kg is on a rough incline that makes an angle of 30° with the ground and it starts to slip up when the other mass is slightly more than 4.0 kg.

• State two differences between static friction and dynamic friction
  2
• Calculate the coefficient of static friction between the mass of 3.0 kg and the rough surface.   
  3
• The thread connecting masses breaks and the mass of 3.0 kg is around 1.5 kg mass and the rough surface is 0.40.   ${ms^{-1}}$
  2
• Determine the speed of the mass of 3.0 m.   ${ms^{-1}}$
  2
  

参考答案:
空格1: 0.96  空格2: 1.5  空格3: 2.4 


本题详细解析:

1. Static friction occurs when the object is not moving (relative to a surface) and Dynamic friction occurs when there is relative motion between the object and a surface $\mathrm{✓}$

   The magnitude of static friction is dependent on external forces and the magnitude of dynamic friction is not dependent on external forces OR static friction force can vary up to a maximum while dynamic friction is a constant value  $\mathrm{✓}$

2. Component of weight parallel to the incline = $\ll mgsin(30\mathrm{°})=(3.0)(9.81)(0.5)=\gg 15\ll N\gg$ and tension in thread $T=mg=(4.0)(9.81)=39N$. $\mathrm{✓}$

   Static friction = $\ll 39-15=\gg 24\ll N\gg$ $\mathrm{✓}$

   coefficient of static friction = $\ll {\displaystyle \frac{24}{(3.0)(9.81)cos(30\mathrm{°})}}=\gg$

   0.96
   $\mathrm{✓}$ (0.94 to 0.96 accepted)

   $\underset{‾}{\text{Further Explanation}}$: 
   According to the diagram, the mass of 3 $kg$ is balanced with weight $W$, tension $T$ static friction $F_f$ and normal force $N$;

   

   Weight has two components parallel to the incline $W_h$ and perpendicular to the incline $W_v$;

   • $W_h=mgsin(30\mathrm{°})=(3.0)(9.81)(0.5)=15N$ down to the slope

   The tension balances the weight of the mass of 4.0 $kg$;

   • $T=mg=(4.0)(9.81)=39N$

   The forces parallel to the incline must cancel each other;

   
   

   

   
   

   $\underline{\text{Further Explanation}}$:There are only two forces that act on the mass parallel to the incline. They are components of the weight 

   $W_h$ down to the slope and dynamic friction $F_f$ up the slope. The dynamic friction;

3. 
   

   

   
   

   

   • $\begin{array}{rl}{\displaystyle W_h+F_f}& {\displaystyle =T}\\ {\displaystyle }\\ {\displaystyle 15+F_f}& {\displaystyle =39}\\ {\displaystyle }\\ {\displaystyle F_f}& {\displaystyle =24N}\end{array}$

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