Question

a 10kg block is set moving wth an initial seped of 6m/s on a rough horizontal surface. If the force of friction is 20 N, approximately how far does the block travel before itstops

Answer 1

Answer:

$x=9m$

Explanation:

According to Newton's second law, we have:

$-F_f=ma(1)$

The force of friction is opposite to the motion of the block.

Using the following kinematic equation, we can calculate the distance traveled by the block before it stops($v_f=0$).

$v_f^2=v_0^2+2ax$

Solving for x and replacing a from (1):

$x=\frac{v_f^2-v_0^2}{2a}\\x=\frac{0-v_0^2}{2(\frac{-F_f}{m})}\\x=\frac{mv_0^2}{2F_f}\\x=\frac{10kg(6\frac{m}{s})^2}{2(20N)}\\x=9m$