Question

A 40 kg slab rests on a frictionless floor as shown in the figure. A 10 kg block rests on the top of the slab. The static coefficient of friction between the block and slab is 0.60 while the kinetic friction is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 m/s², the resulting acceleration of the slab will be.

• A. 1 m/s²
• B. 1.5 m/s²
• C. 2 m/s²
• D. 6 m/s²

Answer 1

The correct option is A

1 m/s²

Step 1. Given data

Mass of the slab, $m_s=40kg$

Mass of the block, $m_b=10kg$

Coefficient of static friction, ${\mu }_s=0.60$

Coefficient of kinetic friction, ${\mu }_k=0.40$

Horizontal force, $F=100newton$

Frictional force $=f$

Acceleration due to gravity, $g=9.8m/s^2$

Step 2. Formula used

$f=\mu mg$

Step 3. Calculation acceleration of block ($a$)

Force of static friction between block and slab, $f_s={\mu }_m\times m_b\times g$

$$\begin{gathered} f_s=0.6\times 10\times 9.8 \\ {f}_s=58.8newton \end{gathered}$$

Since the applied force (100 N) is greater than the force of static friction (58.8 N), therefore the block will slip over the slab.

Due to the movement of the block, kinetic friction will come into play and kinetic friction force ($F_k$ ) will act on the slab.

The free-body diagram of the block-slab interface is shown below.

Force of kinetic friction, $f_k={\mu }_k\times m_b\times g$

$$\begin{gathered} f_k=0.4\times 10\times 9.8 \\ {f}_k=39.2N \end{gathered}$$

Using Newton's second law of motion,$F=ma$

$a=f_k/m_s$

$$\begin{gathered} a=39.2/40 \\ a=0.98 \\ a=1m/s^2 \end{gathered}$$

Hence, option (a) will be correct.