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1. The problem
A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force ##\overrightarrow{F}## shown in the figure. The coefficient of static friction between all surfaces is 0.65 and the kinetic coefficient is 0.37.
If the force is 10% greater than your answer for (a), what is the acceleration of each block?
3. The attempt
For part (a), I correctly calculated the force to be ##F\approx89~\textrm{N}## thus making the new force ##F=97.9 ~\textrm{N}##
Solving for ##a## where ## F_{f3}=## the force of friction between the two blocks and ##F_{f5}=## the force of friction between the lower block and the ground, I get the following:
$$F-T-F_{f3} -F_{f5}=m_5a$$
Substituting ##F_{f3}-m_3a## for Tension:
$$F-2F_{f3}-m_3a-F_{f5}=m_5a\Rightarrow \frac{F-2F_{f3}-F_{f5}}{m_3+m_5}=a\Rightarrow \frac{F-2m_3g\mu-(m_3+m_5) g\mu}{m_3+m_5}=a$$
$$\frac{97.9 \textrm{N} -6\textrm{kg}\cdot 9.8\frac{m}{s^2}(.37)-8\textrm{kg}\cdot9.8\frac{m}{s^2}(.37)}{(8\textrm{kg})}\approx 5.89 \frac{m}{s^2}$$
Is this correct?
If the force is 10% greater than your answer for (a), what is the acceleration of each block?
3. The attempt
For part (a), I correctly calculated the force to be ##F\approx89~\textrm{N}## thus making the new force ##F=97.9 ~\textrm{N}##
Solving for ##a## where ## F_{f3}=## the force of friction between the two blocks and ##F_{f5}=## the force of friction between the lower block and the ground, I get the following:
$$F-T-F_{f3} -F_{f5}=m_5a$$
Substituting ##F_{f3}-m_3a## for Tension:
$$F-2F_{f3}-m_3a-F_{f5}=m_5a\Rightarrow \frac{F-2F_{f3}-F_{f5}}{m_3+m_5}=a\Rightarrow \frac{F-2m_3g\mu-(m_3+m_5) g\mu}{m_3+m_5}=a$$
$$\frac{97.9 \textrm{N} -6\textrm{kg}\cdot 9.8\frac{m}{s^2}(.37)-8\textrm{kg}\cdot9.8\frac{m}{s^2}(.37)}{(8\textrm{kg})}\approx 5.89 \frac{m}{s^2}$$
Is this correct?