Question

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$f_{1}$ and $n_{1}$ appear in both diagrams as action-reaction pairs.

(b) For the $5.00kg$ mass, Newton’s second law in the $y$ direction gives:

$n_{1}=m_{1}g=(5.00kg)(9.80m/s_{2})=49.0N$

In the $x$ direction,

$f_{1}−T=0$

$T=f_{1}=μmg=0.200(5.00kg)(9.80m/s_{2})=9.80N$

For the $10.0kg$ mass, Newton’s second law in the $x$ direction gives:

$45.0N−f_{1}−f_{2}=(10.0kg)a$

In the $y$ direction,

$n_{2}−n_{1}−98.0N=0$

$f_{2}=μn_{2}=μ(n_{1}+98.0N)=0.20(49.0N+98.0N)=29.4N$

$45.0N−9.80N−29.4N=(10.0kg)a$

$a=0.580m/s_{2}$

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