No, not that equation.
final KE=initial PE - frictional loss
1/2 m vf^2=m*g*(12)-7.4E3
solve for vf
A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 71 kg, and the height of the water slide is 12.0 m. If the kinetic frictional force does -7.4 × 103 J of work, how fast is the student going at the bottom of the slide? Use g = 9.81 m/s2
Could I use the equation
vf=sqrt v0^2 + 2(g)(h0-hf) ?
3 answers
so for m
So for m I would plug in 71kg correct?