# Mechanics: Acceleration in Pulley System

1. Sep 16, 2014

### END

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?

2. Sep 17, 2014

### ehild

It is correct. Nice work!

ehild