M Karim Physics Numerical Book Solution Class 11 Online

Using the equation: $$f = \mu N$$, where $\mu$ is the coefficient of friction and $N$ is the normal reaction.

Using Newton's second law of motion: $$F - f = ma$$, where $F$ is the applied force, $f$ is the frictional force, $m$ is the mass, and $a$ is the acceleration.

$$10 = \mu \times 5 \times 9.8$$

$$f = 20 - 10 = 10$$ N

$$\mu = \frac{10}{5 \times 9.8} = 0.2$$

$$a = \frac{20}{5} = 4$$ m/s²

Using the equation of motion: $$v = u + at$$, where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. m karim physics numerical book solution class 11

Given: $F = 20$ N, $m = 5$ kg, $a = 2$ m/s²

$$20 = 0 + a \times 5$$