Question

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 81 kg, and the height of the water slide is 11.7 m. If the kinetic frictional force does -6.7 × 10³ J of work, how fast is the student going at the bottom of the slide?

Answer 1

By law of conservation of energy, the energy is always conserved that is the potential gets converted to kinetic energy.
Hence,
$PE =KE$ where PE is Potential Energy and KE is Kinetic Energy [PE =Mgh]
So,
$KE=Mgh$ = $81 Kg \times 9.8 m/s^2 \times 11.7 m$
$KE =9287.46 J$

Now out of this some Kinetic Energy is lost due to friction.
Hence,
Remaining Kinetic Energy = $9287.46 J - 6700 J$
=$2587.46 J$
So,
Velocity at bottom
Remained KE =$\frac{1}{}2mv^2$

$2587.46 =\frac{1}{}2 \times 81 \times v^2$
Solving, we get
v= 8m/s
So, the student is going at the speed of 8m/s