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 77 kg, and the height of the water slide is 12.9 m. If the kinetic frictional force does -7.1 × 103 J of work, how fast is the student going at the bottom of the slide?

Answer 1

solution :

given data :

mass , m = 77 kg

height , h = 12.9 m

kinetic frictional force = - 7.1 x 10³ J

here , student is initially rest at top of water surface , so student has some potential energy due height from bottom of water surface to top of water surface.

P.E = mgh

= 77 x 9.8 x 12.9

P.E = 9734.34 J

here,when student goes down to bottom of the surface from the top surface then it's potential energy (P.E) is converted in to kinetic energy (K.E) and kinetic frictional force due to friction.

P.E = K.E - kinetic frictional force (due to friction)

{ we deduct this kinrtic frictional force from kinetic energy because it is loss from potential energy )

K.E = P.E + kinetic frictional force (due to friction)

= 9734.34 + ( - 7.1 X 10³ )

= 9734.34 - 7100

K.E = 2634.34 J

kinetic energy of student = K.E = 2634.34 J , when he is goes down to bottom of water surface with speed ( v)

kinetic energy = m v² / 2

2634.34 = 77 x v² / 2

2634.34 x 2 / 77 = v²

5268.68 / 77 = v²

( 68.4244 )^1/2 = v

  v = 8.27 m/s

so here velocity of student is equal to  v = 8.27 m/s when he is goes down to bottom side.