Question

There is a boy with a mass of 55kg starting at rest going down a slide which is 4m high (y distance) and 6m (x distance) and his final velocity at the end is 4.5m/s then what is the average F of friction acting against him as he goes down the slide?

Answer 1

Clearly,while sitting on top of the slide,total energy of the boy is purely potential energy i.e `mgh=55×9.81×4=2158.2J`

Again,when the boy is at the bottom of the slide,his total energy is purely kinetic energy i.e `1/2 ×m×v^2=1/2 × 55 ×(4.5)^2 =556.875 J`

Clearly,during this process of sliding down,`(2158.2 - 556.875)=1601.325J` of energy has been lost,and this has happened because of working against frictional force,

Now,in this case,if frictional force acting is `f`,so for sliding by a distance of `s=sqrt(4^2+6^2)=7.21m`(this is the length of the slide) along the slide,work done against the frictional force is `f*s=f*7.21=1601.325`

So,frictional force acting is `1601.325/7.21 =222.1 N`