Asked by Sarah
                a fun loving 11.4kg otter slides up a hill and then back down to the same place. If she starts up at 5.75 m/s and returns at 3.75 m/s, how much mechanical energy did she lose on the hill and what happened to that energy
            
            
        Answers
                    Answered by
            bobpursley
            
    Figure the KE at start, subtract the KE at the finish, that is the friction loss.
    
                    Answered by
            Sarah
            
    so i did Ki= 1/2mv^2 = 1/2(11.4kg)(5.75 m/s)2 = 188.45 J
Kf= 1/2mv^2
1/2(11.4kg)(3.75m/s)2= 80.15 J
so I took 188.45 - 80.15 and got 108 J is this correct
    
Kf= 1/2mv^2
1/2(11.4kg)(3.75m/s)2= 80.15 J
so I took 188.45 - 80.15 and got 108 J is this correct
                    Answered by
            bobpursley
            
    yes
    
                    Answered by
            Sarah
            
    Thank you! 
    
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