m•g•h- W(fr) = m•v²/2
v = sqrt{2•g•h – [2•W(fr)/m]} =
sqrt{2•9.81•12 – [2•7.4•10³/71]} = 11.44 m/s
1) 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 × 10^3 J of work, how fast is the student going at the bottom of the slide? Use g = 9.81 m/s2
Would I do sqrt (2)(-7.4e^3)+1/2(12)(71) - (71)(9.8)?
1 answer
1 answer