$$a = \frac{20}{5} = 4$$ m/s²
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.
$$\mu = \frac{10}{5 \times 9.8} = 0.2$$
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.
$$f = 20 - 10 = 10$$ N
$$10 = \mu \times 5 \times 9.8$$
A block of mass 5 kg is placed on a horizontal surface. A force of 20 N is applied to the block, causing it to move with a uniform acceleration of 2 m/s². What is the coefficient of friction between the block and the surface? m karim physics numerical book solution class 11
Given: $F = 20$ N, $m = 5$ kg, $a = 2$ m/s²
