Question

The $1.0kg$block in FIGURE EX$7.24$is tied to the wall with a rope. It sits on top of the $2.0kg$block. The lower block is pulled to the right with a tension force of $20N$. The coefficient of kinetic friction at both the lower and upper surfaces of the $2.0kg$block is ${\mu }_k=0.40$

a. What is the tension in the rope attached to the wall?

b. What is the acceleration of the $2.0kg$block?

Short answer

(a) The tension in the rope attached to the wall is $3.92N$

(b) The acceleration of the$2kg$block.

Step-by-step solution

Given information (part a)

Mass of lower block, $M=2kg$

Mass of the upper block, $m=1kg$

External force $F_{ext}=20N$

Coefficient of kinetic friction${\mu }_k=0.4$

Explanation (part a)

The tension in the rope will be equal to frictional force acting on 1 kg mass.

$$\begin{gathered} T={\mu }_{k}mg \\ m=1kg \\ g=9.8m/s^2 \\ T=0.4\left(1kg\right)\left(9.8m/s^2\right) \\ =3.92N \end{gathered}$$

The tension in the rope is$3.92N$

Given information (part b)

Mass of lower block,$M=2kg$

Mass of the upper block,$m=1kg$

External force$F_{ext}=20N$

Coefficient of kinetic friction${\mu }_k=0.4$

Explanation (part b)

Frictional force at the upper and lower surface of 2 kg block will try to prevent its motion.

The equation of motion of 2 kg block is as follows.

$$\begin{gathered} Ma=F_a-T-f_k \\ \text{where}{f}_k={\mu }_k\left(m+M\right)g \\ Ma=F_a-T-{\mu }_k\left(m+M\right)g \\ \left(2kg\right)a=\left(20N\right)-\left(3.92N\right)-\left[\left(0.4\right)\left(1kg+2kg\right)\left(9.8m/s^2\right)\right] \\ \left(2kg\right)a=4.32N \\ a=\frac{4.32N}{2kg} \\ a=2.16m/s^2 \end{gathered}$$

The acceleration of the $2kg$block is $2.16m/s^2$