Q 34E - University Physics with Modern Physics

Question

Question: Consider the system shown in Fig. E5.34. Block A weighs 45.0N, and block B weighs 25.0N. Once block B is set into downward motion, it descends at a constant speed. (a) Calculate the coefficient of kinetic friction between block A and the tabletop. (b) A cat, also of weight 45.0N, falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration (magnitude and direction)?

Step-by-Step Solution

Answer

(a) 0.56

(b) $2.16m/s^2$

Step 1: Identification of given data

Weight of block A is $m_{A}g=45 \text{N}$

Weight of block B is $m_{B}g=25 \text{N}$

Step 2: Significance of Newton’s 2 nd law of motion

Newton's second law states that an object will accelerate in the direction of the net force. This acceleration leads the object to begin moving more slowly before coming to a complete stop since the force of friction acts in the opposite direction from that of motion.

Step 3: (a) Determining the coefficient of kinetic friction between block A and the tabletop

Block B is moving with constant speed so the acceleration will be and the net force also zero, so from free body diagram equate the horizontal and vertical forces

$$\begin{gathered} T-m_{B}g=\sum F_{net} \\ T-m_{B}g=0 \end{gathered}$$

$T=m_{B}g$

For block A

Normal force on block A is $N=m_{A}g$

$$\begin{gathered} T-{\mu }_{K}N=\sum F_{net} \\ T-{\mu }_{K}N=0 \\ T={\mu }_{K}N \end{gathered}$$

From equation (i) 

$\begin{array}{rcl}{\mu }_{K}N& =& m_{B}g\\ {\mu }_K& =& \frac{m_{B}g}{N}\\ & =& \frac{m_{B}g}{m_{A}g}\end{array}$

Substituting all the values in above equation

$\begin{array}{rcl}{\mu }_K& =& \frac{25 \text{N}}{45 \text{N}}\\ & =& 0.56\end{array}$

So the coefficient of kinetic friction between block A and the tabletop is 0.56.

Step 4: (b) Determining the acceleration of block B

A cat weight of , falls asleep on top of block A, so the total weight of block A ($m_{A}g$) will be

$45 \text{N}+45 \text{N}=90 \text{N}$

By applying Newton’s law on block A

$T-{\mu }_{K}N=m_{A}a$                                                                                       …(ii)

Where T tension on rope is, a is acceleration and N is normal force

By applying Newton’s law on block B

$m_{B}g-T=m_{B}a$                                                                                       …(iii)

From equation (ii) and equation (iii)

$$\begin{gathered} m_{B}g-({\mu }_{K}N+m_{A}a)=m_{B}a \\ {m}_{B}a+m_{A}a=m_{B}g-{\mu }_{K}N \\ a=\frac{m_{B}g-{\mu }_K{m}_{A}g}{m_B+m_A} \end{gathered}$$

Substitute all the values in above equation

$\begin{array}{rcl}a& =& \frac{25 \text{N}-0.56\times 90 \text{N}}{\frac{25 \text{N}}{g}+\frac{90\text{N}}{g}}\\ & =& \frac{-25.4g}{115}\\ & =& -2.16 {\text{m/s}}^{\text{2}}\end{array}$

Hence the acceleration of block B is $-2.16m/s^2$ (-ve sign indicate that block B is moving upward direction).