Space Time & Motion A.2 Forces & Momentum (id: ecd50d3a4)

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 Firsf ahbispeuk5nsj+tilp+(,k 24-.ko.t Law.p*3iaaqtp+0qbrn8+ s A block of mass 2.0 kg slides at the constant speed of v on a rough inclined plane. A string attached to the hanging mass of 0.50 kg 2.passes over a pulley and supportsbaip+pr +nta q s*38q0 the block. The slope of the incline is 25°.


[Source: Created with Chemix - https://chemix.org/]
1.Show that the direction of the motion of the box is down the slope.
2.Draw and label a free-body diagram for the forces acting on the block.

3.Determine the coefficient of dynamic friction between the block and the inclined plane.  
4.In reality, air resistance will also have an effect on the motion of the block and the calculated coefficient of friction. If air resistance were taken into account for this constant speed scenario, discuss how the calculated value would be affected.

参考答案: 0.19


本题详细解析:
1.When the net force acti v/bjv+;wte+ d*:6k faa3nq nmng on the body is zero

 The body has constant velocity/does not accelerate An object at constant velocity ≪and in a state of rest≫ will remain so Unless acted on by an external/unbalanced force 

2.1 Tension/Weight of the mass hunger=(0.5)(9.81)=4.9N 

The component of the weight of the block on the plane of incline(2.0)(9.81)(sin25)=(2.0)(9.81)(0.42) = 8.3N 8.3>4.9

 Further Explanation:

 The direction of the friction on the block should be in the opposite direction of the block's motion. 

There is no acceleration. According to Newton's second law, there is no net force either on the object separately or the system. 

The forces on the block must be balanced. 

3. The net force acting on the block should be zero. The forces on the plane of incline are the force of dynamic friction, the tension up to the slope and the component of weight down to the slope. The force of dynamic friction $F_f$;

   $F_f={\mu }_d{F}_N$

   The component of the weight of the block $W_{Tx}$ on the plane of the incline;

   $W_{Tx}=mgsin25\mathrm{°}=(2.0)(9.81)(0.42)=8.3N$

   The tension $T$;

   $T=W_m=mg=(0.50)(9.81)=4.9N$

   The forces that are perpendicular to the plane of the incline are normal reaction force and the component of the weight of the block.

   The component of the weight of the block $W_{Ty}$ perpendicular to the plane of the incline;

   $W_{Ty}=mgcos25\mathrm{°}=(2.0)(9.81)(0.91)=18N$

   $W_{Ty}$ must be equal to $R$.

   So, dynamic friction can be rewritten as;

   $F_f={\mu }_d{W}_{Ty}$

   Since $W_{Tx}$ must be balanced with $F_f$ and $T$;

   $F_f+T=W_{Tx}$

   
   

   ${\mu_d}{W_{Ty}}$=${W_{Tx}}-T$

   $\mu_d$=$\frac{{W_{Tx}}-T}{W_{Ty}}$=$\frac{8.9-4.9}{18}$=0.19

   
   

   4.air resistance also contributes a force up the slope, which means there is less force due to dynamic friction.

   the calculated value for the coefficient of friction would be lower 

   Further Explanation:

   The direction of air resistance is always opposite to the motion of the object. Therefore, it is in the same direction as the force of dynamic friction and tension. Instead of the force of dynamic friction with tension only, the resultant force of air friction, tension and the force of dynamic friction balances the component of weight. Hence, the force of dynamic friction would be lower. This would lead to a lower calculated value for coefficient of friction

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