Question

A boy of mass 80 kg slides down a vertical pole, and a frictional force of 480 N acts on him. What is his acceleration as he slides down the pole? (Use g = 10 m/ss)

Answer 1

Answer:

His acceleration is $\overrightarrow{a}=4\frac{m}{s^{2}}$

Explanation:

Newton's second law states that acceleration of a body is cause by a net force, the relation between them is:

$\sum\overrightarrow{F}=m\overrightarrow{a}$

On the boy there're acting two forces, his weight (W) that points downward and the frictional force (f) that points upward (they boy moves downward and friction always is opposite to movement). So $\sum\overrightarrow{F}=\overrightarrow{W}+\overrightarrow{f}$ so (1) is:

$\overrightarrow{W}+\overrightarrow{f}=m\overrightarrow{a}$

Using the positive direction downward weight and gravitational acceleration(g) are positive and friction force is negative:

$W-f=m\overrightarrow{a}$, solving for a:

$\overrightarrow{a}=\frac{W-f}{m}$, weight is mg:

$\overrightarrow{a}=\frac{mg-f}{m}=\overrightarrow{f}=\frac{(80kg)(10\frac{m}{s^{2}})-(480)}{80}$

$\overrightarrow{a}=4\frac{m}{s^{2}}$