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

`x=9m`

Step-by-step 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=(v_f^2-v_0^2)/(2a)\\x=(0-v_0^2)/(2((-F_f)/(m)))\\x=(mv_0^2)/(2F_f)\\x=(10kg(6(m)/(s))^2)/(2(20N))\\x=9m`