Space Time & Motion A.2 Forces & Momentum (id: cfef26c86)

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]
A box of mass 3kg is ph56nd1f (o) brj+l ww1y.whlml+l0mbb3cp4 x+(e fdja (/js ktaced on a box of mass 7kg. The coefficient of static friction between the surfaces of the boxes is 0.5. A horizontal forceF is applied to the 7kg box which sits on a f(tpcx3dk(ja jbs b +e4 +lm0/frictionless surface.

What is the maximum value of F that can move the boxes without the 3kg box slipping?

A. 0.1g
B. 0.5g
C. 1g
D. 5g


参考答案:  D


本题详细解析:

$\underline{\text{Explanation}}$:

The free-body diagrams of the boxes:

The maximum value of the force of friction $F_f$ can be calculated as follows;

• $F_f={\mu }_s{F}_N={\mu }_{s}mg=(0.5)(3)g=1.5g$

When Newton's Second Law is applied to the mass of $3kg$;

• $F_{net}=ma$

• $1.5g=3a$

• $0.5g=a$

To move without slipping, both of the masses must have the same acceleration. Treating both boxes as one system

• $F_{net}=ma$

• $F_{net}=(10)(0.5g)$

• $F=5g$

Hence, the correct answer is D.

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