A 20 kg child on roller skates, initially at rest, rolls 4 m down an incline at an angle of 30 degrees with the horizontal. If there is no friction between incline and skates, what is the kenetic energy of the child at the bottom of the incline? (g= 9.81 m/s^2)

Answer 1

There are two ways of doing this problem, the energy way and the work way.
The energy way is to say the potential energy lost by moving down is the kinetic energy gained at the bottom.
Potential energy lost = m g times the change in height
The change in height is from:
sin (30) = change in height / 4
so change in height = 4 sin 30 = 2 meters down
so the potential energy lost is:
m * 9.8 * 2 which is 19.6 m Joules
That 19.6 m Joules is the kinetic energy at the bottom
KE = 19.6 m
since m = 20 kg
KE = 20*19.6 = 392 Joules
NOW the other way
The work done is the force down the ramp times the distance run
The component of force down the ramp is
m g cos 60 = 20 *9.8 * .5 = 98 Newtons
That 98 newtons pushes a distance of 4 meters so the work done is
98 * 4 = 392 Joules
That work done is used to increase the kinetic energy of the child so the kinetic energy is also 392 Joules.