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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

1 Answer

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Answer:


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

User Solarissmoke
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