Answer:
[tex]10.27m/s[/tex]
Explanation:
Given data
work W= -5.10 10^3 J
mass m= 71kg
final height of slide h2= 12.7m
initial height of slide h1=0m
initial velocity v1= 0m/s
final velocity v2=?
Step two:
required
Final velocity
The work-energy theorem is expressed as'
[tex]W=1/mv_2^2 +mgh_2-(1/mv_1^2+mgh_1)[/tex]
make V2 subject of formula we have final speed
[tex]v_2=\sqrt{\frac{2W}{m}+v_1^2-2g(h_1-h_2) } \\\\[/tex]
substitute our given data we have
[tex]v_2=\sqrt{\frac{2*(-5.1*10^3)}{71}+0^2-2*9.81(12.7) } \\\\v_2=\sqrt{143.66-249.174 } \\\\v_2=\sqrt{105.514 } \\\\v_2=10.27m/s[/tex]
The student going at 10.27m/s
