Space Time & Motion A.2 Forces & Momentum (id: d5358ec52)

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

[单选题]
Block A of mass $m_A$=3.0 kg $m_A$=3.0kg rests above the flat horizontal surface of block B of mass $m_B$=6.0kg, which in turn rests on a smooth horizontal surface as shown.

The coefficients of static and kinetic friction between A and B are $\mu_s$​=0.2 and $\mu_k$​=0.1 respectively. A horizontal force of F=7.0N is applied towards the right on block A. Assuming acceleration due to gravity g≈10$ms^{-2}$ the direction and magnitude of acceleration $a_B$of blockB are:

A. To the left. 1.0$ms^{-2}$
B. To the left 0.5$ms^{-2}$
C. To the right 1.0$ms^{-2}$
D. To the right 0.5$ms^{-2}$


参考答案:  D


本题详细解析:

$\underline{\text{Explanation}}$: 
The normal contact force on $A$, $R_A=m_{A}g=(3.0kg)(10m{s}^{-2})=30N$

(Contact force from $B$ supports the weight of $A$).

The maximum friction on between $A$ and $B$ is $F_s={\mu }_s{R}_A=(0.2)(30N)=6.0N$.

The applied force $F=7.0N>f_{s,max}=6.0N$.

Therefore, block $A$ slides on block $B$, reducing the final frictional force between the blocks to kinetic friction, $F_k={\mu }_k{R}_A=0.1\times 30N=3.0N$.

On $A$, this friction will be to the left (as it is being pulled to the right).

Since friction on $B$ is a reaction to friction on $A$, it will be to the right, and equal to $3.0N$. (Newton's third law).

$a_B={\displaystyle \frac{F_{net}}{m_B}}={\displaystyle \frac{f_k}{m_B}}={\displaystyle \frac{3.0N}{6.0kg}}=0.5m$

微信扫一扫，分享更方便