Question

A $10kg$ block is placed on a horizontal surface. The coefficient of friction between them is $0.2$ . A horizontal force P=15 N first acts on it in the eastward direction. Later, in addition to P a second horizontal force Q=20 N acts on it in the northward direction. Choose the correct option of the following.

• A. The block will not move when only P acts, but will move when both P and Q act.
• B. If the block moves, its acceleration will be 0.5 m/s${}^2$.
• C. When the block moves, its direction of motion will be $ta{n}^{-1}\left(\frac{4}{3}\right)$ east of north.
• D. When both P and Q act, the direction of the force of friction acting on the block will be $ta{n}^{-1}\left(\frac{3}{4}\right)$ west of south.

Answer 1

Answer: (A), (B), (D)

The block will not move when only P acts, but will move when both P and Q act.
If the block moves, its acceleration will be 0.5 m/s${}^2$.
When both P and Q act, the direction of the force of friction acting on the block will be $ta{n}^{-1}\left(\frac{3}{4}\right)$ west of south.

$Maximumfriction=\mu mg=0.2\times 10\times 10=20N$

When P starts acting, frictional force of 15N starts acting westward and block does not move. When Q starts acting:

$F=\sqrt{{15}^2+{20}^2}=25N$

Direction of this force is at an angle with north:

$\theta ={\tan }^{-1}\left(\frac{15}{20}\right)={\tan }^{-1}\left(\frac{3}{4}\right)$

Direction of frictional force is opposite to this.

Thus, frictional force at ${\tan }^{-1}\left(\frac{3}{4}\right)$ west of south.

$accelerationofblock=\frac{F-f_s}{10}=0.5m/s^2$

Thus, options A, B and D are correct.