$$20 - f = 5 \times 2$$
Given: $F = 20$ N, $m = 5$ kg, $a = 2$ m/s² m karim physics numerical book solution class 11
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? $$20 - f = 5 \times 2$$ Given:
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. Using Newton's second law of motion: $$F -
$$f = 20 - 10 = 10$$ N
Using the equation: $$f = \mu N$$, where $\mu$ is the coefficient of friction and $N$ is the normal reaction.
$$10 = \mu \times 5 \times 9.8$$