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## Homework Statement

A 0.75-kg block slides on a frictionless surface with a speed of 20 m/s. It then slides over a rough area 1.0m in length and onto another frictionless surface. The coefficient of kinetic friction between the block and the rough surface is 0.17.What is the speed of the block after it passes across the rough surface?

m = 0.75 kg

x(during friction being applied)= 1.0m

Coefficient of Friction = 0.17

Initial Velocity = 20 m/s

Gravity = 9.8

x(during friction being applied)= 1.0m

Coefficient of Friction = 0.17

Initial Velocity = 20 m/s

Gravity = 9.8

## Homework Equations

-Equations I used -

() - means multiplying.

Normal Force = Mass(Gravity)

*-Just used the F*

_{x}component since F_{y}equals 0-=

Friction Force= Coefficient(Normal Force)(Distance

_{of the friction})

K = 1/2mv

^{2}

E

_{i}= E

_{f}

=

K

_{i}- Friction Force = K

_{f}

=

1/2mv

^{2}-Coefficient(Normal Force)(Distance)= 1/2mv

^{2}

## The Attempt at a Solution

**I first realized what I needed to find, which was the Final Velocity.**

**Then set up my equation.**

1/2mv

^{2}-Coefficient(Normal Force)(Distance)= 1/2mv

^{2}

**Then plugged in the numbers.**

1/2(.75

_{kg})(20

_{m/s})-(.17

_{f})(.75

_{kg})(9.8

_{m/s2})(1.0

_{m}) = 1/2(.75

_{kg})(v

^{2})

**Simplified.**

6.2505 = 3.75 v

^{2}

**Simplified again.**

1.6668 = v

^{2}

**Squared.**

1.291 = v

**Books Answer**

**However, When I looked at the book they got the answer 0.82m/s and got it by doing...**

v=√(InitialV

^{2}) - 2 (Coefficient of friction)(Gravity)(Distance))

** Note - all of this is under the square root **

v = √(2.0)

^{2}-2(0.17)(9.8)(1.0)

=0.82 m/s

My question is... What did I do wrong and what did the book do correctly? I would really love a good explanation on this as It's important for me to grasp this concept. The answer is there but I want to know HOW and WHY. Thank you in advance whoever takes the time to answer this it is greatly appreciated and your positivity will be rewarded.

Kind regards, A lost soul.